Answer
9.0 L
Work Step by Step
Moles of Al= $\frac{mass\,of\,Al}{molar\,mass\,of\,Al}=\frac{6.5\,g}{26.98\,g/mol}$
$=0.2409 mol$
When 2 moles of Al is reacted, according to the equation, 3 moles of $H_{2}$ is produced.
1 mol of Al produces $\frac{3}{2}=1.5$ moles of $H_{2}$.
$\implies$ moles of $H_{2}$ produced (n)$=1.5\times0.2409\,mol=0.36135\,mol$
$P=742\, mmHg\times\frac{1\,atm}{760\, mmHg}=0.9763\,atm$
$T=(22.0+273)K= 295\,K$
$PV=nRT$ (ideal gas law)
$\implies V= \frac{nRT}{P}=\frac{(0.36135\,mol)(0.0821\,L\,atm\,mol^{-1}K^{-1})(295\,K)}{0.9763\,atm}$
$=9.0\,L\,H_{2}$
