Question

Iron(III) oxide reacts with carbon to give iron and carbon monoxide:Fe2O3(s)+3C(s)→2Fe(s)+3CO(g)How many grams of C are required to react with 24.3 g of Fe2O3 ?

Answer 1

5.48 grams of C

Step-by-step explanation

Given:

Mass of Fe₂O₃ that reacted = 24.3 g

Equation: Fe₂O₃(s) + 3C(s) → 2Fe(s) + 3CO(g)

What to find:

The grams of C that are required to react with 24.3 g of Fe₂O₃(s).

Step-by-step solution:

According to the equation; 1 mole of Fe₂O₃(s) reacts with 3 moles of C

1 mole of Fe₂O₃ = 159.69 g

1 mole of C = 12.011 g, so 3 moles of C = 3 x 12.011 g = 36.033 g

Therefore, the mass of C required to react with 24.3 g of Fe₂O₃(s) is calculated as follows:

`\begin{gathered} 159.69\text{ }g\text{ }Fe₂O₃=36.033\text{ }g\text{ }C \\ \\ 24.3\text{ }g\text{ }Fe₂O₃=x \\ \\ x=\frac{24.3\text{ }g\text{ }Fe₂O₃}{159.69\text{ }g\text{ }Fe₂O₃}*36.033\text{ }g\text{ }C \\ \\ x=5.48\text{ }g\text{ }of\text{ }C \end{gathered}`

Thus, 5.48 grams of C are required to react with 24.3 g of Fe₂O₃(s).