3. consider the following reaction to answer the questions below: 3fe(s) + 4h₂o(g) → fe₃o₄(s) + 4h₂(g). a…

Explanation:

Step1: Calculate moles of reactants

Molar mass of $H_2O$ is $18.02\ g/mol$, molar mass of $Fe$ is $55.85\ g/mol$. Moles of $H_2O=\frac{10.00\ g}{18.02\ g/mol}\approx0.555\ mol$ Moles of $Fe = \frac{18.00\ g}{55.85\ g/mol}\approx0.322\ mol$

Step2: Use mole - ratio from balanced equation

The mole - ratio of $Fe$ to $H_2O$ in the reaction $3Fe(s)+4H_2O(g)\rightarrow Fe_3O_4(s)+4H_2(g)$ is $\frac{3}{4}$. For $0.555\ mol$ of $H_2O$, moles of $Fe$ required $=0.555\ mol\times\frac{3}{4}= 0.416\ mol$ Since we have only $0.322\ mol$ of $Fe$, $Fe$ is the limiting reactant.

Step3: Calculate moles of $H_2O$ used

Moles of $H_2O$ used based on the amount of $Fe$: Since the mole - ratio of $Fe$ to $H_2O$ is $\frac{3}{4}$, moles of $H_2O$ used $=0.322\ mol\times\frac{4}{3}\approx0.429\ mol$

Step4: Calculate moles of excess reactant remaining

Moles of $H_2O$ remaining $=0.555\ mol - 0.429\ mol = 0.126\ mol$

Step5: Calculate mass of excess reactant remaining

Mass of $H_2O$ remaining $=0.126\ mol\times18.02\ g/mol\approx2.27\ g$

Answer:

a. The limiting reactant is $Fe$. b. The mass of the excess reactant ($H_2O$) remaining unused is approximately $2.27\ g$.