Answer
2.31 mol $H_{2}O$
Work Step by Step
Balanced chemical reaction:
$C_{2}H_{5}OH$ (l) + 3 $O_{2}$g ${\longrightarrow}$ 2$CO_{2}$(g)+3$H_{2}O$(l)
1 mol of $C_{2}H_{5}OH$ gives 3 moles of 3$H_{2}O$(l). So the conversion factor for 0.77 mol $C_{2}H_{5}OH$ to water would be: $\frac{3_{mol (H_{2}O)}}{1_{mol (C_{2}H_{5}OH)}}$.
Therefore the amount of moles $H_{2}O$ produced by 0.77 mol $C_{2}H_{5}OH$ is:
0.77 mol $C_{2}H_{5}OH$ x $\frac{3_{mol (H_{2}O)}}{1_{mol (C_{2}H_{5}OH)}}$ = 2.31 mol $H_{2}O$
