$2$ $Al(s)$ + $3$ $H_2SO_4(aq) to $ $Al_2(SO_4)_3(aq)$ + $3$ $H_2(g)$
First calculate the number of moles of $Al$ in $7.25$ $g$, knowing the molar mass of $Al$ is $27$ $gmol^-1$:
$n=m/M=7.25/27=0.27$ $mol$
From the equation we can see that each $2$ $mol$ of $Al$ yields $3$ $mol$ of $H_2$, so multiply this by $3/2$ to find the number of moles of $H_2$ released.
$n=0.40$ $mol$
We know that a mole of an ideal gas occupies $22.4$ $L$ at standard temperature and pressure (STP), which is $0^o$ $C$ or $273$ $K$ and $1$ $atm$, so this number of moles will produce $0.4xx22.4=9.0$ $L$ of gas at STP.
We are not asked about STP, though, but $25^o$ $C$ ($273$ $K$) and $1.65$ $atm$.
Using the :
$(P_1V_1)/T_1=(P_2V_2)/T_2$
Rearranging to make $V_2$ the subject:
$V_2=(P_1V_1)T_2/T_1P_2=(1*9*298)/(273*1.65)=5.95$ $L$
