A substance contains $37.5\% \; C$. $12.5\% \; H$ and $50.0\% \; O$.
Calculate the volume $CO_2$ (under normal conditions) and the mass of water that one collects on burning $10\;g $
$m_C $ in $10\; g $ = $3.75\; g $
$n_C$ in $10\; g $ = $\frac{3.75}{12} = 0.3125 $
$m_H $ in $10\; g $ = $1.25\; g $
$n_H$ in $10\; g $ = $\frac{1.25}{12}$ $=$ $0.1042 $
$n_{CO_2}$ obtained = $n_C$
$V_{CO_2}$ NTP obtained= $n_{CO_2} \cdot 22.4$ $=$ $7.0\; L$
$n_{H_2O}$ obtained = $\frac{n_H}{2}$
$m_{H_2O}$ obtained = $n_{H_2O} \cdot 18$ $=$ $0.938\; g$
