Answer
$Fe$: $15,000\text{ kg}$
$H_2SO_4$: $26,816\text{ kg}$
Work Step by Step
Required $H_2$ to fill $4800m^{3}$ is $=\frac{4800}{0.80}=6000 m^{3}$
Temperature $T=(0+273)=273K$
Pressure $P=1atm$
Then $n=\frac{PV}{RT}=\frac{1 \times 6000\times 10^{3}}{0.082\times 273}\approx 268\times 10^{3} mol$
$Fe(s)+H_2SO_4(aq)=FeSO_4(aq)+H_2(g)$
Here
$1mol Fe$ reacts with $1mol H_2SO_4$ to produce $1mol H_2$
So the required mass of $Fe$ is:
$268\times 10^3\times 55.85\approx 15,000\text{ kg}$
and the mass of $H_2SO_4$:
$260\times 10^{3} mol\times 98.08\approx 26,280\text{ kg}$
Since the acid is $98\%$ by mass, the total solution is:
$\frac{26,280}{0.98}=26,816\text{ kg}$
