Answer:
1.25 Liters of 6.0 m solution sulfuric acid is required.
Explanation:
[tex]2Al(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2(g)[/tex]
Moles of hydrogen gas = [tex]\frac{15.0 g}{2 g/mol}= 7.5 mol[/tex]
According to reaction, 3 moles of hydrogen gas are obtained from 3 moles of sulfuric acid.
Then 7.5 moles of hydrogen will be obtained from:
[tex]\frac{3}{3}\times 7.5 mol=7.5 mol[/tex]
Moles of Sulfuric acid : 7.5 mol
Molarity of the sulfuric acid solution = 6.0 M
Volume of the solution = V
[tex]6.0 M=6.0 mol/L=\frac{7.5 mol}{V}[/tex]
[tex]V=\frac{7.5 mol}{6.0 mol/L}=1.25 L[/tex]
