Chemistry High School

In one experiment, 7.62 g of iron are allowed to react with 8.67 g of sulfur. Calculate the mass of iron (Ill) sulfide that can be formed.

1) What’s the Mass of Fe2S3?
2) What is the limiting reactant?

Answers

Answer 1

Answer:

The mass of iron sulphide formed is 14.2 g

Explanation:

Starting off with a balanced chemical equation:

2Fe + 3S --> Fe₂S₃

to find the mass of Fe₂S₃ and thus the limiting reagent, we can first find the number of moles of each reactant. Number of moles is found by dividing mass by molar mass (found using a standard IUPAC Periodic Table).

n(Fe) present = m/M = 7.62/55.85 = 0.13644 mol

n(S) present = m/M = 8.67/32.07 = 0.27035 mol.

Using the values for the number of moles of each reactant, we can find the limiting reagent by testing each with the molar ratio of the equation.

If (Fe) is the limiting reagent, then moles of sulphur required = n(Fe)/2 × 3

= 0.20466 mol. Since we have excess of what's required of sulphur, then iron could be the limiting reagent.

Now we test it with (S). If (S) is the limiting reagent, then moles of iron required = n(S)/3 × 2 = 0.18023 mol. Since we have less than what is required, therefore, sulphur is NOT the limiting reagent.

Hence, Iron is the limiting reagent.

Now we have the limiting reagent, we can use this to calculate the number of moles of Fe₂S₃, since the limiting reagent is the reactant that is completely consumed.

Since stoichiometry of Fe : Fe₂S₃ = 2 : 1, therefore:

moles of (Fe) = 2× moles of (Fe₂S₃).

hence, moles of (Fe₂S₃) = 1/2 × 0.13644 mol = 0.06822.

Finally, we can multiply the moles of iron sulphide by the molar mass, to get the mass.

m(Fe₂S₃) = nM = 0.06822 × (55.85×2 + 32.07×3) = 14.2 g (3 significant figures, since all the data in the question is in 3 sig. figs).

Related Questions

2. What is the topic of the question?
Which substance forms propanoic acid as one of the products when it reacts with hot
concentrated acidified potassium manganate(VII)?
A but-1-ene
B but-2-ene
C 2-methylpropene
D 2-methylbut-1-ene mel
Why did you choose this answer?
nouzou
no
oorlo voy

Answers

The potassium permanganate is an inorganic compound with the chemical formula KMnO₄. Its IUPAC name is given as potassium manganate (VII). The  substance which forms propanoic acid is but-1-ene. The correct option is A.

What is oxidation?

The oxidation is defined as the process which involves the addition of oxygen or removal of hydrogen. The reaction of potassium manganate (VII) with an alkene is an oxidation reaction.

The acidified potassium manganate (VII) solution oxidizes the alkene by breaking the 'C'-'C' double bond and replacing it with 'C'-'O' double bonds.

The reaction of but-1-ene in presence of potassium manganate (VII) is:

CH₃CH₂CH = CH₂ → CH₃CH₂COOH + CO₂

Thus the correct option is A.

To know more about oxidation, visit;

https://brainly.com/question/29218241

#SPJ1

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