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Chapter 9: Problem 135

Iron(III) oxide reacts with carbon monoxide to give iron metal and carbon dioxide. If you begin the reaction with \(24.0 \mathrm{~g}\) of iron(III) oxide and \(34.0 \mathrm{~g}\) of carbon monoxide, what is the theoretical yield in grams of carbon dioxide?

Short Answer

Expert verified
The theoretical yield of carbon dioxide for the reaction of iron(III) oxide with carbon monoxide, given 24.0 g of iron(III) oxide and 34.0 g of carbon monoxide, is \(19.8 \mathrm{~g}\).

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the reaction of Iron(III) oxide with carbon monoxide is: \(Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\)
02

Calculate the moles of reactants

First, we need to find the molar mass of each compound: \(Fe_2O_3\): (2 * 55.85) + (3 * 16.00) = 159.70 g/mol \(CO\): 12.01 + 16.00 = 28.01 g/mol Next, we'll calculate the moles of each reactant: moles of \(Fe_2O_3 = \frac{24.0 \mathrm{~g}}{159.70 \mathrm{~g/mol}} = 0.15 \mathrm{~mol}\) moles of \(CO = \frac{34.0 \mathrm{~g}}{28.01 \mathrm{~g/mol}} = 1.21 \mathrm{~mol}\)
03

Determine the limiting reactant

Using the balanced equation and the moles we calculated, we'll find the moles needed for each reactant to completely react: moles of \(CO\) needed for complete reaction with \(Fe_2O_3: 0.15 \mathrm{~mol} \times \frac{3 \mathrm{~mol \ CO}}{1 \mathrm{~mol \ Fe_2O_3}} = 0.45 \mathrm{~mol \ CO}\) Since we have more than enough moles of \(CO\) (1.21 mol) compared to the 0.45 mol required for a complete reaction, iron(III) oxide is the limiting reactant.
04

Calculate the moles of carbon dioxide produced

Using the moles of the limiting reactant (iron(III) oxide) and the stoichiometry from the balanced equation, we can find the moles of \(CO_2\) produced: moles of \(CO_2 = 0.15 \mathrm{~mol \ Fe_2O_3} \times \frac{3 \mathrm{~mol \ CO_2}}{1 \mathrm{~mol \ Fe_2O_3}} = 0.45 \mathrm{~mol \ CO_2}\)
05

Convert the moles of carbon dioxide to grams

Finally, we need to calculate the mass of carbon dioxide produced. The molar mass of \(CO_2\) is: \(CO_2\): 12.01 + (2 * 16.00) = 44.01 g/mol Multiply the moles of \(CO_2\) by its molar mass to get the theoretical yield in grams: theoretical yield of \(CO_2 = 0.45 \mathrm{~mol} \times 44.01 \mathrm{~g/mol} = 19.8 \mathrm{~g}\) The theoretical yield of carbon dioxide for this reaction is 19.8 grams.

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