how many grams of CO are needed to react with an excess of Fe2O3 to produce 209.7gFe? Fe2O3(s) + 3CO(g) => 3CO2(g) + 2Fe(s)
Almost the same procedure as the last problem you posted. After you have mols, convert to grams by grams = mols x molar mass.
Post your work if you get stuck and explain in detail why you don't know what to do next.
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To solve this problem, you first need to determine the balanced chemical equation for the reaction. The given equation is already balanced:
Fe2O3(s) + 3CO(g) => 3CO2(g) + 2Fe(s)
Next, we need to calculate the molar mass of Fe, CO, and Fe2O3, which are:
Molar mass of Fe (Fe atomic mass): 55.85 g/mol
Molar mass of CO (C atomic mass + O atomic mass):
12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Molar mass of Fe2O3 (Fe atomic mass + O atomic mass):
(2 * 55.85 g/mol) + (3 * 16.00 g/mol) = 159.69 g/mol
Next, we need to determine the ratio between Fe2O3 and Fe by comparing their coefficients in the balanced equation. From the equation:
2 moles of Fe are produced from 1 mole of Fe2O3.
Now, we can begin the calculation. The steps are as follows:
Step 1: Calculate the number of moles of Fe.
Moles of Fe = mass of Fe / molar mass of Fe
Moles of Fe = 209.7 g / 55.85 g/mol ≈ 3.75 mol
Step 2: Determine the number of moles of Fe2O3 based on the mole ratio.
From the balanced equation, we know that 1 mole of Fe2O3 produces 2 moles of Fe.
Moles of Fe2O3 = (Moles of Fe) / 2
Moles of Fe2O3 = 3.75 mol / 2 ≈ 1.875 mol
Step 3: Calculate the mass of Fe2O3 using the number of moles and molar mass.
Mass of Fe2O3 = moles of Fe2O3 * molar mass of Fe2O3
Mass of Fe2O3 = 1.875 mol * 159.69 g/mol ≈ 299.59 g
Therefore, approximately 299.59 grams of CO are needed to react with an excess of Fe2O3 to produce 209.7 grams of Fe.