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Answer:To determine how many grams of oxygen are used to produce 35.0 g of water, we need to use stoichiometry based on the given chemical equation:
\[ \text{C}_7\text{H}_{16} + 11\text{O}_2 \rightarrow 8\text{H}_2\text{O} + 7\text{CO}_2 \]
First, let's follow these steps:
1. **Calculate the moles of water (H\(_2\)O) produced:**
\[ \text{Molar mass of H}_2\text{O} = 2(1.01 \, \text{g/mol}) + 16.00 \, \text{g/mol} = 18.02 \, \text{g/mol} \]
\[ \text{Moles of H}_2\text{O} = \frac{35.0 \, \text{g}}{18.02 \, \text{g/mol}} \approx 1.942 \, \text{mol} \]
2. **Determine the moles of O\(_2\) used:**
According to the balanced chemical equation, 8 moles of H\(_2\)O are produced for every 11 moles of O\(_2\):
\[ \frac{11 \, \text{mol O}_2}{8 \, \text{mol H}_2\text{O}} = x \, \text{mol O}_2 \]
\[ x = 1.942 \, \text{mol H}_2\text{O} \times \frac{11 \, \text{mol O}_2}{8 \, \text{mol H}_2\text{O}} \approx 2.670 \, \text{mol O}_2 \]
3. **Convert moles of O\(_2\) to grams:**
\[ \text{Molar mass of O}_2 = 2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol} \]
\[ \text{Grams of O}_2 = 2.670 \, \text{mol} \times 32.00 \, \text{g/mol} = 85.44 \, \text{g} \]
Therefore, to produce 35.0 g of water, you used approximately 85.44 grams of oxygen.
Explanation:
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