Question #fae95

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Question #fae95

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Answer 1 · 2017-02-13T03:10:33

$1040 g$ $1040g$

Explanation:

The balanced equation for the combustion of octane is:
$2 C_{8} H_{18} + 25 O_{2} \to 16 C O_{2} + 18 H_{2} O$ $2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O$

Therefore,
a. Calculate the number of moles of octane:
$235 g \cdot \frac{1mol}{114.26g} = 2.96 m o l$ $235g * "1mol"/"114.26g" = 2.96mol$

b. Calculate the grams of carbon dioxide:
$2.96 mol octane \cdot \frac{16 mol CO2}{2 mol octane} \cdot \frac{44.01 g C O 2}{1 m o l C O 2} = 1042 g$ $2.96 "mol octane" * (16 "mol CO2")/(2 "mol octane") * (44.01g CO2)/(1 mol CO2) = 1042g$

(To the appropriate number of sig figs: 1040g).

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Question #fae95

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