Determine the volume of sulfuric acid solution needed to pre

Determine the volume of sulfuric acid solution needed to prepare 37.4 g of aluminum sulfate, Al2(SO4)3, by the following reaction

2Al(s) +3H2SO4 (aq) -> Al2(SO4)3 (aq) + 3H2 (g)

The sulfuric acid solution, whose density is 1.104 g/ml, contains 15.0 % H2SO4 by mass

37.4g aluminium sulphate = 37.4/342.15=0.11moles

1 mole aluminium sulphate is given by 3 moles H2SO4

so 0.11 moles aluminium sulphate is given by 0.33 moles H2SO4

0.33 moles H2SO4 = 0.33*98 = 32.34gm

as acid solution is 15% H2SO4 wt of acid required = 32.34/0.15 = 215.6 gm

so volume = mass/density = 215.6/1.104 = 195.289ml.....................................ANS