Answer
Water formed during combustion of 26.7 g butane =$ 41.4 g$
Work Step by Step
$2 C_{4}H_{10} + 13H_{2}O → 8CO_{2} + 10 H_{2}O$
The above equation is the balanced equation for the combustion of butane.
Molecular mass of butane $C_{4}H_{10} = 4 \times 12 + 10 \times1$ = 58 g
According to stoichiometry we can write,
116 g butane yields 180 g water.
Therefore water formed during combustion of 26.7 g butane =
$ (26.7 \times180) \div 116 = 41.4 g$
