Iron(III) oxide reacts with carbon to give iron and carbon

Step 1 of 3:

(a)  Here, we are going to find out the amount of C are required to react with 16.5 g of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\)

We know that

\(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \rightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(g)\)

The molar mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}=159.69 \mathrm{~g} / \mathrm{mol}\)

Given mass = 16.5 g

Thus, the amount of \(\mathrm{Fe}_{2} \mathrm{O}_{3}=16.5 \mathrm{~g} \times 1 / 159.69 \mathrm{~g} / \mathrm{mol}=0.103 \text { moles }\)

We know,

\(\begin{array}{l} \frac{1 \text { mole of } \mathrm{Fe}_{2} \mathrm{O}_{3}}{3 \text { moles of } \mathrm{C}}=\frac{0.103 \text { mole of } \mathrm{Fe}_{2} \mathrm{O}_{3}}{n \text { moles of } \mathrm{C}} \\ \Rightarrow n=0.3099 \text { moles of } \mathrm{C} \end{array}\)

The molar mass of C = 12.01 g/mol

Thus, the amount of C are required to react with 16.5 g of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\)

\(=12.01 \mathrm{~g} / \mathrm{mol} \times 0.3099 \text { moles }=3.72 \mathrm{~g}\)