The amount of carbon dioxide evolved upon complete combustion of $116$ g of $n$-butane is
(Given: atomic mass in amu H $=1$, C $=12$ and O $=16$)

The balanced chemical equation for the combustion of $n$-butane is:

$C_4H_{10} + \dfrac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O$

Molar mass of $n$-butane ($C_4H_{10}$) $= 4 \times 12 + 10 \times 1 = 58$ g/mol

Number of moles of $n$-butane $= \dfrac{116}{58} = 2$ moles

From the balanced equation, $1$ mole of $n$-butane produces $4$ moles of $CO_2$.

Number of moles of $CO_2$ produced $= 2 \times 4 = 8$ moles

Molar mass of $CO_2 = 12 + 2 \times 16 = 44$ g/mol

Mass of $CO_2$ produced $= 8 \times 44 = 352$ g

Answer: $352$ g