Chapter U4, Problem 20STP

Interpretation Introduction

Interpretation:

The correct option for the limiting reactant and the number of moles of Fe₃O₄ formed in the given reaction needs to be identified when 16.8 g Fe is reacted 10.0 g of H₂O.

Concept introduction:

A chemical equation is representation of a chemical reaction in the form of formulae and symbols of molecules and compounds involved. Here, the reactants are present on the left side of the reaction arrow and the products are on the right-hand side.

Limiting reactant is defined as a reactant which limits the product formed in the reaction. This reactant actually decides the yield of the reaction.

Answer to Problem 20STP

The option (A) is correct.

Explanation of Solution

Below is the balance chemical equation:

${\text{3Fe(s) + 4H}}_{\text{2}}\text{O(g) }\stackrel{\Delta }{\to }{\text{ Fe}}_{\text{3}}{\text{O}}_{\text{4}}{\text{(s) + 4H}}_{\text{2}}\text{O(g)}$

Since,

Molar mass of Fe = 55.845 g/mol

Molar mass of H₂O = 18 g/mol

Molar mass of Fe₃O₄ = 231.533 g/mol

16.8 g Fe

10.0 g of H₂O

Here 3 moles of Fe and 4 moles of H₂O combines to form 1 mole of Fe₃O₄.

Now, the amount of product formed from Fe and H₂O is calculated as follows:

From Fe:

$\left(16.8\text{ g Fe}\right)\left(\frac{\text{1 mol Fe}}{55.85\text{ g Fe}}\right)\left(\frac{1{\text{ mol Fe}}_{\text{3}}{\text{O}}_{\text{4}}}{3\text{ mol Fe}}\right)=0.100{\text{ mol Fe}}_{\text{3}}{\text{O}}_{\text{4}}$

From H₂O:

$\left(10.8{\text{ g H}}_2\text{O}\right)\left(\frac{{\text{1 mol H}}_2\text{O}}{18.02{\text{ g H}}_2\text{O}}\right)\left(\frac{1{\text{ mol Fe}}_{\text{3}}{\text{O}}_{\text{4}}}{4{\text{ mol H}}_2\text{O}}\right)=0.139{\text{ mol Fe}}_{\text{3}}{\text{O}}_{\text{4}}$

Here, limiting reagent is Fe since it yields less Fe₃O₄ compare to H₂O. The H₂O is in excess. The yield of Fe₃O₄ is 0.100 mol.

Therefore, option A is correct.