:Answer : The limiting reactant is and the theoretical yield of methanol is, 0.96 grams.
Explanation :
First we have to calculate the moles of and .
where,
= pressure of CO gas = 232 mmHg = 0.305 atm (1 atm = 760 mmHg)
V = volume of gas = 1.65 L
T = temperature of gas = 305 K
= number of moles of CO gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
and,
where,
= pressure of gas = 374 mmHg = 0.492 atm (1 atm = 760 mmHg)
V = volume of gas = 1.65 L
T = temperature of gas = 305 K
= number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of
So, 0.0601 moles of react with moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 2 mole of react to give 1 mole of
So, 0.0601 moles of react with moles of
Now we have to calculate the mass of
Therefore, the theoretical yield of methanol is, 0.96 grams.
The theoretical yield of methanol is 0.496 g of methanol.
The reaction equation is CO (g) + 2H2 (g) → CH3OH (g).
From the partial pressures of each reactant, we can obtain the number of moles of reactants.
For CO;
P = 232 mmHg or 0.305 atm
V = 1.5 L
T = 305 K
n = ?
R = 0.082 atmL-1mol-1K-1
PV = nRT
n = PV/RT
n = 0.305 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K
n = 0.018 moles
For hydrogen;
P = 397 mmHg or 0.522 atm
V = 1.5 L
T = 305 K
n = ?
R = 0.082 atmL-1mol-1K-1
PV = nRT
n = PV/RT
n = 0.522 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K
n = 0.031 moles
From the reaction equation;
1 mole of CO reacted with 2 moles of H2
0.018 moles of CO will react with 0.018 moles × 2 moles/1 mole
= 0.036 moles of H2
We can see that there is not enough H2 to react with CO hence H2 is the limiting reactant.
2 moles of H2 yields 1 mole of methanol
0.031 moles of H2 yields 0.031 moles × 1 moles/2 mole
= 0.0155 moles of methanol
Mass of methanol produced = 0.0155 moles of methanol × 32 g/mol
= 0.496 g of methanol
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Example: One liter of saturated calcium fluoride
solution contains 0.0167 gram of CaFat 25°C.
Calculate the molar solubility of, and Ksp for, CaF2.
Answer:
[tex]Molar\ solubility=2.14x10^{-4}M[/tex]
[tex]Ksp=3.91x10^{-11}[/tex]
Explanation:
Hello,
In this case, given that 0.0167 grams of calcium fluoride in 1 L of solution form a saturated one, we can notice it is the solubility, therefore, the molar solubility is computed by using the molar mass of calcium fluoride (78.1 g/mol):
[tex]Molar\ solubility=\frac{0.0167gCaF_2}{1L}*\frac{1molCaF_2}{78.1gCaF_2} \\\\Molar\ solubility=2.14x10^{-4}M[/tex]
Next, since dissociation equation for calcium fluoride is:
[tex]CaF_2(s)\rightarrow Ca^{2+}(aq)+2F^-(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]
We can compute the solubility product by remembering that the concentration of both calcium and fluoride ions equals the molar solubility, thereby:
[tex]Ksp=(2.14x10^{-4})(2*2.14x10^{-4})^2\\\\Ksp=3.91x10^{-11}[/tex]
Regards.
1. If a question states "The patching material costs NASA $306/in2 …", the conversion factor in this statement is: Recognizing — Equation statement: $306 = _____ ___________
Answer:
$306 = Cost of 1 square inch of the patching material in question.
$306 = 1 in²
Explanation:
The conversion factor is am expression that is used to prove the equivalence of some quantities with different units.
The conversion factor basically converts from one quantity to another.
For this question, the conversion factor given for the patching material is $306/in².
This means that the patching material costs $306 for every square inch, the equation for the conversion is thus
$306 = 1 in² of the patching material.
Hope this Helps!!!
Please help me out ASAP!
What is meant by concentration?
Answer:
concentration is the abundance of a constituent divided by the total volume of a mixture.
Which of the following is not part of the proper protocol for using acids and bases?
A. Storing acids and bases below 10°C
B. Add acid to water, not water to acid
C. Storing acids and bases in separate areas
D. Wearing protective clothing while handling
Answer:
A. Storing acids and bases below 10C
Explanation:
I took the exam and got it correct :)
Storing acids and bases below 10°C is not part of the proper protocol for using acids and bases.
What is an acid?
Acids are defined as substances which on dissociation yield H+ ions , and these substances are sour in taste.Compounds such as HCl, H₂SO₄ and HNO₃ are acids as they yield H+ ions on dissociation.
According to the number of H+ ions which are generated on dissociation acids are classified as mono-protic , di-protic ,tri-protic and polyprotic acids depending on the number of protons which are liberated on dissociation.
Acids are widely used in industries for production of fertilizers, detergents batteries and dyes.They are used in chemical industries for production of chemical compounds like salts which are produced by neutralization reactions.
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Benzene boils at 80.10 °C and has a molal boiling constant, k b, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 °C. What is the molality of the solute?
Answer:
[tex]m=0.395mol/kg[/tex]
Explanation:
Hello,
This is a problem about boiling point elevation which is modeled via:
[tex]\Delta T=i*m*Kb[/tex]
Whereas for this solvent (nonpolar, nonionizing), the van't Hoff factor is one. In such a way, the molality of the solute is simply computed as shown below:
[tex]m=\frac{\Delta T}{Kb}=\frac{(81.10-80.10)\°C}{2.53\°C/m} \\\\m=0.395mol/kg[/tex]
In this manner, we can also compute the molar mass of the solute by noticing 20.0 g (0.020 kg) of benzene were used:
[tex]n=0.395mol/kg*0.020kg=7.9x10^{-3} mol[/tex]
And considering the 2.15 g of the solute:
[tex]Molar\ mass=\frac{2.15g}{7.9x10^{-3}mol}\\ \\Molar\ mass=271.975g/mol[/tex]
Best regards.
Calculate the equilibrium concentrations of N2O4 and NO2 at 25 ∘C in a vessel that contains an initial N2O4 concentration of 0.0655 M . The equilibrium constant Kc for the reaction N2O4(g)⇌2NO2(g) is 4.64×10−3 at 25 ∘C. Express your answers using four decimal places separated by a comma.
Answer:
[N2O4] = 0.0573M
[NO2] = 0.0163M
Explanation:
The equilibrium of N2O4 is:
N2O4(g)⇌2NO2(g)
Where Kc is defined as:
Kc = 4.64x10⁻³ = [NO2]² / [N2O4]
When you add just N2O4, the reaction will occurs until [NO2]² / [N2O4] = 4.64x10⁻³. Here, the system reaches equilibrium.
That means if 0.0655M N2O4 begin reaction, in equilibrium we will have:
[N2O4] = 0.0655M - X
[NO2] = 2X
Where X is defined as reaction coordinate
Replacing in Kc:
4.64x10⁻³ = [NO2]² / [N2O4]
4.64x10⁻³ = [2X]² / [0.0655-X]
3.0392x10⁻⁴ - 4.64x10⁻³X = 4X²
3.0392x10⁻⁴ - 4.64x10⁻³X - 4X² = 0
Solving for X:
X = -0.0093 → False solution. there is no negative concentrations
X = 0.008156M → Right solution.
Replacing X, equilibrium concentrations are:
[N2O4] = 0.0655M - X
[NO2] = 2X
[N2O4] = 0.0573M[NO2] = 0.0163MIdentify some other substances (besides KCl) that might give a positive test for chloride upon addition of AgNO3. Based on the reactant used in your experiment, do you think it is reasonable to exclude these types of substances as contaminants that would give a false positive when you tested your reaction residue to verify that it is KCl?
Answer:
Other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide. However iodides and bromides have different colours hence they will not give a false positive test for KCl. Other chlorides present may lead to a false positive test for KCl.
Explanation:
In the qualitative determination of halogen ions, silver nitrate solution is used. Various halide ions give various colours of precipitate with silver nitrate. Chlorides yield a white precipitate, bromides yield a cream precipitate while iodides yield a yellow precipitate. All these ions or some of them may be present in the system.
However, if other chlorides are present, they will also yield a white precipitate just as KCl leading to a false positive test for KCl. Since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. We can exclude other halides from the tendency to lead us to a false positive test for KCl but not other chlorides.
In the qualitative analysis test of chloride upon addition of AgNO₃, presence of diffrerent chloride salts other than KCl will gives false test.
How do we get on addition of KCl in AgNO₃?On adding halogens on the silver nitrate solution we will get the precipitate of diffrent colors of diffrent halides.
Chlorides will gives white color precipiate in the silver nitrate solution, so KCl and other chlorides will also give white color precipitate.Bromides will gives cream color precipitate.Iodides will gives yellow color precipitate.So, presence of diffrent chloride salt in the silver nitrate solution in addition with KCl will gives a false positive result for the test.
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Question 14 of 25
What type of reaction is BaCl2 + Na,504 → 2NaCl + Baso,?
A. Single-replacement
B. Synthesis
C. Double-replacement
D. Decomposition
double displacement
bcoz each of the reactants combines with other reactants to obtain the product
Identify each reaction from the citric acid cycle as an oxidation‑reduction reaction, an esterification reaction, an amidation reaction, a hydrolysis reaction, a hydration reaction, or a dehydration reaction.
1. Which type of reaction occurs when succinyl-CoA is converted to succinate in the citric acid cycle?
2. Which type of reaction occurs when malate is converted to oxaloacetate in the citric acid cycle?
3. Which type of reaction occurs when aconitate is converted to isocitrate in the citric acid cycle?
Answer:
1. Oxidation-reduction and hydrolysis
2. Oxidation-reduction
3. Dehydration
Explanation:
Our options for each reaction are:
a) Oxidation‑reduction reaction
b) Esterification reaction
c) Amidation reaction
d) Hydrolysis reaction
c) Hydration reaction
f) Dehydration reaction
In reaction one the have the rupture of the S-CoA bond. This reaction takes place by the addition of a water molecule and the oxidation to a carboxylic acid group. So, for reaction 1 we will have an oxidation-reduction and a hydrolysis reaction.
For reaction 2, the functional group change from alcohol to a carboxylic acid. So, we have an oxidation-reduction reaction.
In the last reaction, we have the production of a double bond by the removal of water. With this in mind, we have a dehydration reaction.
See figure 1
I hope it helps
how are mass and weight affected in chemical reactions?
Answer:
How the chemical reacts
Explanation:
How to do this
Q1 and Q2
Only want to know how to find molecular formula
Answer:
Question 1
A. Empirical formula is C8H8O3
B. Molecular formula is C8H8O3
Question 2.
A. Empirical formula is CH2
B. Molecular formula is C4H8
Explanation:
Question 1:
A. Determination of the empirical formula:
Carbon (C) = 63.2%
Hydrogen (H) = 5.26%
Oxygen (O) = 31.6%
Divide by their molar mass
C = 63.2/12 = 5.27
H = 5.26/1 = 5.26
O = 31.6/16 = 1.975
Divide by the smallest
C = 5.27/1.975 = 2.7
H = 5.26/1.975 = 2.7
O = 1.975/1.975 = 1
Multiply through by 3 to express in whole number
C = 2.7 x 3 = 8
H = 2.7 x 3 = 8
O = 1 x 3 = 3
Therefore, the empirical formula for the compound is C8H8O3
B. Determination of the molecular formula of the compound.
From Avogadro's hypothesis, 1 mole of any substance contains 6.02×10²³ molecules.
Now from the question given, we were told that 1 molecule of the compound has a mass of 2.53×10¯²² g.
Therefore, 6.02×10²³ molecules will have a mass of = 6.02×10²³ x 2.53×10¯²² = 152.306 g
Therefore, 1 mole of the compound = 152.306 g
The molecular formula of the compound can be obtained as follow:
[C8H8O3]n = 152.306
[(12x8) + (1x8) + (16x3)]n = 152.306
[(96 + 8 + 48 ]n = 152.306
152n = 152.306
Divide both side by 152
n = 152.306/152
n = 1
The molecular formula => [C8H8O3]n
=> [C8H8O3]1
=> C8H8O3
Question 2:
A. Determination of the empirical formula of the compound.
Mass sample of compound = 0.648 g
Carbon (C) = 0.556 g
Mass of Hydrogen (H) = mass sample of compound – mass of carbon
Mass of Hydrogen (H) = 0.648 – 0.556
Mass of Hydrogen (H) = 0.092 g
Thus, the empirical formula can be obtained as follow:
C = 0.556 g
H = 0.092 g
Divide by their molar mass
C = 0.556/12 = 0.046
H = 0.092/1 = 0.092
Divide by the smallest
C = 0.046/0.046 = 1
H = 0.092/0.046 = 2
Therefore, the empirical formula of the compound is CH2.
B. Determination of the molecular formula of the compound.
Mole of compound = 0.5 mole
Mass of compound = 28.5 g
Molar mass of compound =.?
Mole = mass /Molar mass
0.5 = 28.5/ Molar mass
Cross multiply
0.5 x molar mass = 28.5
Divide both side by 0.5
Molar mass = 28.5/0.5 = 57 g/mol
Thus, the molecular formula of compound can be obtained as follow:
[CH2]n = 57
[12 + (1x2)]n = 57
14n = 57
Divide both side by 14
n = 57/14
n = 4
Molecular formula => [CH2]n
=> [CH2]4
=> C4H8.
g If attempting to dissolve both silver bromide and silver chloride in aqueous solution through complex ion formation, which data will be the most relevant?
Answer:
Kf
Explanation:
The stability constant Kf of a given complex specie is an equilibrium constant that represents the formation of that particular complex specie in solution. It measures the strength of the interaction between the ligands and metal that form the particular complex specie. The magnitude of Kf shows how easily a complex specie is formed in solution.
Hence if I want to dissolve the bromides or chlorides of silver which are ordinarily insoluble in water by means of complex formation, the magnitude of the stability constant for each particular complex specie is important as it gives information regarding the thermodynamic feasibility of the process.
A plot of 1/[BrO-] vs. time is linear and the slope is equal to 0.056 M-1s-1. If the initial concentration of BrO- is 0.65 M, how long will it take one-half of the BrO- ion to react
Answer:
time taken for one-half of the BrO⁻ ion to react is t= 27.45 secs
Explanation:
equation of reaction
3BrO⁻(aq) → BrO₃⁻(aq) + 2Br⁻(aq) (second order reaction)
given
the slope of the graph is 0.056M⁻¹s⁻¹ = k(constant)
initial concentration [A]₀ = 0.65M
for second order reaction,we can calculate the time taken for one-half of the BrO- ion to react using:
[tex]\frac{1}{[A]}[/tex] =[tex]\frac{1}{[A]}[/tex]₀ ⁺ k × t
where initial concentration [A]₀ = 0.65M
[A] = [A]₀÷2 = 0.325M
[tex]\frac{1}{0.325M}[/tex] = [tex]\frac{1}{0.65M}[/tex] + 0.056M⁻¹s⁻¹ × t
3.077= 1.54 + 0.056t
3.077-1.54=0.056t
1.537=0.056t
t= 27.45 secs
1. What is the frequency of light with a wavelength of 1064 nm?
2. How many photons of 1064 nm light are contained in a 535 Joules of this light?
Answer:
1) 2.8×10^14 Hz
2) 2.88×10^21 photons
Explanation:
Recall that the formula for the speed of a wave is v=λf, since we are talking about light, we can replace v with c hence; c=λf. Where;
c= speed of light = 3 ×10^8 m/s
λ= wavelength of light= 1064nm= 1064×10^-9 m
f= frequency of light= the unknown
Hence;
f= c/λ= 3×10^8/1064×10^-9
f= 2.8×10^14 Hz
Energy of a single photon=hf= 6.626×10^-34 Js × 2.8×10^14 s^-1 = 18.55×10^-20J
If 1 photon contains 18.55×10^-20J of energy
x photons contains 535 J of energy
x= 535/18.55×10^-20J
x= 2.88×10^21 photons
oxygen get stable configuration by ____________two electrons
please give the answer as fast as you can
please
Answer:
gaining two electrons
Explanation:
electron configuration
2:6
so add two to 6 to get stable 2:8
Iron(II) is available to bond with chloride ion. How many of each type of ion will bond to form an ionic compound?
A) 3 iron(II), 1 chloride
B) 2 iron(II), 3 chloride
C) 2 iron(II), 1 chloride
D) 1 iron(II), 2 chloride
Answer:
D) 1 iron(II), 2 chloride
Explanation:
Iron II chloride is the compound; FeCl2. It is formed as follows, ionically;
Fe^2+(aq) + 2Cl^-(aq) -----> FeCl2
The formation of one mole of FeCl2 involves the reaction one mole of iron and two moles of chloride ions. This means that in FeCl2, the ratio of iron to chlorine is 1:2 as seen above.
Therefore there is one iron II ion and two chloride ions in each mole of iron II chloride, hence the answer.
Potassium iodide reacts with lead(II) nitrate in the following precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) What minimum volume of 0.400 M potassium iodide solution is required to completely precipitate all of the lead in 310.0 mL of a 0.112 M lead(II) nitrate solution?
Answer:
0.1736 L or 173.6 ml
Explanation:
Number of moles of lead II nitrate is obtained by;
Number of moles = concentration × volume of solution
Concentration= 0.112 M
Volume of solution= 310 ml
n= 0.112 × 310/1000
n= 0.03472 moles
From the reaction equation;
2 moles of potassium iodide reacted with 1 mole of lead II nitrate
x moles of potassium iodide will react with 0.03472 moles of lead II nitrate
x= 2 × 0.03472 moles= 0.06944 moles of potassium iodide
Volume of potassium iodide solution = number of moles/ concentration = 0.06944/ 0.4
Volume of potassium iodide solution= 0.1736 L or 173.6 ml
Which best describes the total mass of a sample of water when it condenses
from a liquid to a gas?
A. The mass is less because the water molecules get closer together
and take up more space.
B. The mass is the same because the decrease in energy equals the
increase in the number of molecules.
C. The mass is the same because water molecules are not created or
destroyed during a phase change.
D. The mass is greater after water condenses because the mass of
the molecules increases.
Answer:
Its C I hopefully help you
Which of the following statements is not true for an exothermic reaction? Question options: The products have a higher heat content than the reactants. The temperature of the reaction system increases. The temperature of the surroundings increases. Heat passes from the reaction system to the surroundings. The enthalpy change for the reaction is negativ
Answer:
The products have a higher heat content than the reactants.
Explanation:
The statement above is not true for an exothermic reaction because in an exothermic reaction heat is released to the surroundings. This simply means that the total energy of the products is less than that of the reactants.
14. Which group of diamagnetic transition metals exhibits trends in density and melting points that don't match the same trends seen in
other groups?
A. Group 3
B. Group 12
C. Group 7
D. Group 11
Answer:
Group 12
Explanation:
Group 12 transition metals are diamagnetic. They behave properties that distinguish them. They naturally have twelve electrons hence their outermost shell is fully filled.
Transition metals have high densities which increases down the group. However, the increase in density of transition elements of group 12 varies with temperature at a rate that is quite different from other transition elements. Hence the differences in the value of melting points and density changes by only a very small amount as you come down group 12 compared to other groups of transition elements.
Need help finding major products
Answer:
Explanation:
RX + AgNO₃ = R⁺ ( carbocation ) + AgX + NO₃⁻
C₂H₅OH ( a nucleophile ) + R⁺ = ROC₂H₅
C₅H₁₁X + AgNO₃ = C₅H₁₁⁺ + AgX + NO₃⁻
In the first case carbocation produced is CH₃CH₂CH₂CH₂CH₂⁺
CH₃CH₂CH₂CH₂CH₂⁺ ⇒ CH₃CH₂CH₂C⁺HCH₃ ( secondary carbocation more stable )
CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃
Hence option D is correct .
b )
In the second case carbocation produced is
CH₃CH₂CH₂CH⁺CH₃
CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃
The product formed is same as in case of first
Option B is correct
Reduction occurs at which electrode?
Answer:
negative charge electrode
Explanation:
In cathode positive ions are picked up to perform reduction.At the same time negative ions are picked up at anode to get oxidized from electrolyte.
Answer:
The electrode that removes ions from the solution :) a p e x
Differentiate between
expansion of solid and liquid
In a liquid, the expansion is a little more than in
solids. The bonds in a liquid are weaker than in a
solid, so as you heat up a liquid, the particles can
move around each other faster and in so doing,
move further apart. Solids and liquids occupy a
'set' volume at a certain temperature.
8.670 mL + 9.87 mL=
Answer:18.4 ML
Explanation:
easy add
A. Identify the structure drawn below.
Answer:
Hexane
Explanation:
You have a carbon structure with only single bonds. This means that the name will end in -ane.
There are 6 carbon atoms. This means that the name will begin with hex-.
The structure is hexane.
Consider the reaction between two solutions, X and Y, to produce substance Z: aX + bY → cZ When 500. mL of a 1.8 M solution of X is combined with 500. mL of a 1.8 M solution of Y, the resulting solution has a concentration of 0.60 M Y and 0.60 M Z. No more of substance X remains in the flask. 1. How many moles each of X and Y are present before the reaction occurs? 2. How many moles each of Y and Z are present after the reaction occurs? 3. How many moles each of X and Y have reacted? 4. What is the balanced equation for this reaction?
Answer:
1. 0.90 are the initial moles of X and Y
2. 0.60 moles are the moles of Y and Z after the reaction
3. 0.90 moles of X and 0.30 moles of Y
4. 3X + 1Y → 2Z
Explanation:
1. For the reaction, initial moles of X and Y are:
500mL = 0.500L × (1.8 moles / L) = 0.90 are the initial moles of X and Y
2. After the reaction. The total volume is 500mL + 500mL = 1L
Moles Y and Z = 1L × (0.60 moles / 1L) = 0.60 moles are the moles of Y and Z after the reaction
3. As there is no moles of X after the reaction, all X reacts, that is 0.90 moles of X. And moles of Y that reacts are 0.90 mol - 0.60mol = 0.30 moles of Y
4. That means 3 moles of X reacts per mole of Y 0.90/0.30 = 3. Also, 2 moles of Z are produced per mole of Y 0.60/0.30 = 2.
That means balanced equation is:
aX + bY → cZ
3X + 1Y → 2ZProvide the structures of the fragments that result when the molecular ion of 2-heptanone undergoes fragmentation via McLafferty rearrangement. Include charges and single electrons.
Answer:
See explanation
Explanation:
We have to start, remembering the mechanism behind the McLafferty rearrangement. The hydrogen in the gamma carbon (in this case, carbon 5) would be removed by a heterolytic rupture due to the cation-radical placed in the oxygen of the carbonyl group. Then we will have several heterolytic ruptures. Between carbons alpha and beta (in this case, 4 and 3) and a rupture in the carbonyl group. Due to these ruptures, two double bonds would be formed. One double bond in the alcohol cation-radical and the other one in the alkene.
See figure 1
I hope it helps!
what is the meaning of the word tetraquark?
Answer:
A tetraquark in physics is an exotic meson composed of four valence quarks.
Explanation:
It has been suspected to be allowed by quantum chromodynamics, the modern story of strong interactions.
Hope it helps.
How do protons and neutrons stay together in the nucleus
Explanation:
The nucleus of an atom is held together by the strong nuclear force that binds together protons and neutrons. Although the strong nuclear force is the strongest of the four fundamental forces, it acts only over very short - typically nuclear - distances. It binds together the protons and neutrons in the nucleus.
Calculate how much acetylene (C2H2) will be produced from 358 g of H2O and an excess of CaC2 if the percent yield for this reaction is 94.5%. CaC2 2 H2O --> C2H2 Ca(OH)2
Answer:
244.7 g of acetylene
Explanation:
The balanced reaction equation is shown below;
2H20 (l) + CaC2 (s) → Ca(OH)2 (s) + C2H2 (g)
Number of moles of were reacted = reacting mass/molar mass = 358g/18gmol-1 = 19.89 moles of water
From the balanced reaction equation;
2 moles water yields 1 mole of acetylene
19.89 moles of water will yield 19.89 × 1/2 = 9.945 moles of acetylene
Theoretical yield of acetylene = 9.945 moles of acetylene × molar mass of acetylene
Molar mass of acetylene = 26.04 g/mol
Theoretical yield of acetylene = 9.945 moles of acetylene × 26.04 g/mol
Theoretical yield of acetylene = 258.9678 g of acetylene
% yield = actual yield/ theoretical yield × 100
94.5 = actual yield/258.9678 g × 100
Actual yield= 94.5 × 258.9678 g/100
Actual yield = 244.7 g of acetylene