Question

How many grams of C are required to react with 97.1 g of Fe2O3?

Answer 1

To find the mass of carbon required to react with 97.1 g of Fe2O3, convert the mass of Fe2O3 to moles, then use the stoichiometric ratio from the balanced equation to find the moles of carbon needed, and finally convert that to grams of carbon. The result is 21.9 g of carbon.

Step-by-step explanation:

The question asks for the mass of carbon (C) needed to react with a given mass of iron(III) oxide (Fe2O3). Although the chemical equation provided is for the reaction of Fe2O3 with carbon monoxide (CO), we will assume a similar reaction involving carbon directly, which is a common reaction in metallurgy:

Fe2O3 + 3C → 2Fe + 3CO

To solve this, we follow a sequence of steps. First, we convert the given mass of Fe2O3 to moles using the molar mass of Fe2O3 (159.7 g/mol). Then, we use the stoichiometric ratio from the balanced equation to find the moles of carbon required. Finally, we convert moles of carbon back to grams using the molar mass of carbon (12.01 g/mol).

Convert 97.1 g of Fe2O3 to moles: 97.1 g / 159.7 g/mol = 0.608 moles of Fe2O3

According to the equation, 1 mole of Fe2O3 reacts with 3 moles of C, so 0.608 moles of Fe2O3 will require 3 × 0.608 moles of C = 1.824 moles of C

Now convert moles of C to grams: 1.824 moles × 12.01 g/mol = 21.9 g of C

The mass of carbon required to react with 97.1 g of Fe2O3 is 21.9 grams, rounded to three significant figures.

Answer 2

21.9 g C

Step-by-step explanation:

Fe₂O₃ + 3 C ⇒ 2 Fe + 3 CO

This is your chemical equation.

To find the grams of C required, first convert grams of Fe₂O₃ to moles using the molar mass (159.69 g/mol).

(97.1 g)/(159.69 g/mol) = 0.608 mol Fe₂O₃

Then, use the mole ratio between Fe₂O₃ and C to convert moles of Fe₂O₃ to moles of C. You can find the mole ratio by looking at the chemical equation.

(0.608 mol Fe₂O₃) × (3 mol Fe₂O₃/1 mol Fe₂O₃) = 1.824 mol C

You can then convert moles of C to grams using the molar mass (12.01 g/mol).

(1.824 mol) × (12.01 g/mol) = 21.9 g