QuestionJuly 20, 2025

If 588 grams of FeS_(2) is allowed to react with 352 grams of O_(2) according to the following unbalanced equation, how many grams of Fe_(2)O_(3) are produced? FeS_(2)+O_(2)arrow Fe_(2)O_(3)+SO_(2) A 4268 B 3208 C 8808 D 3528 E 1108

If 588 grams of FeS_(2) is allowed to react with 352 grams of O_(2) according to the following unbalanced equation, how many grams of Fe_(2)O_(3) are produced? FeS_(2)+O_(2)arrow Fe_(2)O_(3)+SO_(2) A 4268 B 3208 C 8808 D 3528 E 1108
If 588 grams of FeS_(2) is allowed to react with 352 grams of O_(2) according to the following unbalanced equation, how many grams of Fe_(2)O_(3) are produced? FeS_(2)+O_(2)arrow Fe_(2)O_(3)+SO_(2) A 4268 B 3208 C 8808 D 3528 E 1108

Solution
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Answer

E 1108 Explanation 1. Balance the chemical equation 4FeS_{2} + 11O_{2} \rightarrow 2Fe_{2}O_{3} + 8SO_{2} 2. Calculate molar masses Molar mass of FeS_{2} = 119.98 g/mol, O_{2} = 32.00 g/mol, Fe_{2}O_{3} = 159.69 g/mol. 3. Determine moles of reactants Moles of FeS_{2} = \frac{588}{119.98} \approx 4.90 mol. Moles of O_{2} = \frac{352}{32.00} = 11.00 mol. 4. Identify limiting reactant From balanced equation, 4 mol FeS_{2} requires 11 mol O_{2}. FeS_{2} is limiting because 4.90 mol FeS_{2} would require 13.475 mol O_{2}, but only 11.00 mol O_{2} is available. 5. Calculate moles of Fe_{2}O_{3} produced From balanced equation, 4 mol FeS_{2} produces 2 mol Fe_{2}O_{3}. Moles of Fe_{2}O_{3} = \frac{4.90}{4} \times 2 = 2.45 mol. 6. Convert moles of Fe_{2}O_{3} to grams Grams of Fe_{2}O_{3} = 2.45 \times 159.69 \approx 391.27 g.

Explanation

1. Balance the chemical equation<br /> $4FeS_{2} + 11O_{2} \rightarrow 2Fe_{2}O_{3} + 8SO_{2}$<br /><br />2. Calculate molar masses<br /> Molar mass of $FeS_{2}$ = 119.98 g/mol, $O_{2}$ = 32.00 g/mol, $Fe_{2}O_{3}$ = 159.69 g/mol.<br /><br />3. Determine moles of reactants<br /> Moles of $FeS_{2}$ = $\frac{588}{119.98} \approx 4.90$ mol.<br /> Moles of $O_{2}$ = $\frac{352}{32.00} = 11.00$ mol.<br /><br />4. Identify limiting reactant<br /> From balanced equation, 4 mol $FeS_{2}$ requires 11 mol $O_{2}$.<br /> $FeS_{2}$ is limiting because 4.90 mol $FeS_{2}$ would require 13.475 mol $O_{2}$, but only 11.00 mol $O_{2}$ is available.<br /><br />5. Calculate moles of $Fe_{2}O_{3}$ produced<br /> From balanced equation, 4 mol $FeS_{2}$ produces 2 mol $Fe_{2}O_{3}$.<br /> Moles of $Fe_{2}O_{3}$ = $\frac{4.90}{4} \times 2 = 2.45$ mol.<br /><br />6. Convert moles of $Fe_{2}O_{3}$ to grams<br /> Grams of $Fe_{2}O_{3}$ = $2.45 \times 159.69 \approx 391.27$ g.
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