To determine the moles of \(CO_2\) and \(H_2O\) produced from the complete combustion of ethanol (\(C_2H_5OH\)), we can use the balanced chemical equation:
\[ C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O \]
From the equation, we can see the following stoichiometric ratios:
- 1 mole of ethanol produces 2 moles of carbon dioxide (\(CO_2\))
- 1 mole of ethanol produces 3 moles of water (\(H_2O\))
Given that we have 5.0 moles of ethanol, we can calculate the moles of \(CO_2\) and \(H_2O\) produced as follows:
Moles of \(CO_2\) Produced
From the equation: \[ 1 \text{ mole of } C_2H_5OH \rightarrow 2 \text{ moles of } CO_2 \]
Thus, for 5.0 moles of ethanol: \[ 5.0 \text{ moles of } C_2H_5OH \rightarrow 5.0 \times 2 = 10.0 \text{ moles of } CO_2 \]
Moles of \(H_2O\) Produced
From the equation: \[ 1 \text{ mole of } C_2H_5OH \rightarrow 3 \text{ moles of } H_2O \]
Thus, for 5.0 moles of ethanol: \[ 5.0 \text{ moles of } C_2H_5OH \rightarrow 5.0 \times 3 = 15.0 \text{ moles of } H_2O \]
Summary
- Moles of \(CO_2\) produced: 10.0 moles
- Moles of \(H_2O\) produced: 15.0 moles
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