The combustion of n-butane ($C_4H_{10}$) in oxygen is represented by the following balanced equation: $$2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O$$
Molar mass of $C_4H_{10} = (4 \times 12) + (10 \times 1) = 48 + 10 = 58 \text{ g/mol}$.
Given mass = 116 g. $$\text{Moles of } C_4H_{10} = \frac{116}{58} = 2 \text{ moles}$$
From the balanced equation, 2 moles of $C_4H_{10}$ produce 8 moles of $CO_2$. Since we have exactly 2 moles of $C_4H_{10}$, we will produce 8 moles of $CO_2$.
Molar mass of $CO_2 = 12 + (2 \times 16) = 44 \text{ g/mol}$. $$\text{Mass of } CO_2 = 8 \text{ moles} \times 44 \text{ g/mol} = 352 \text{ g}$$
Final Answer: 352 g
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