What is the minimum amount of 6.0 M H2SO4 necessary to produce 15.0 g of H2(g) according to the reaction:

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

Answer :

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

#### Explanation:

Since you are given grams of H2 it will be easiest to work backwards using the stoichiometry.

Therefore:

15g H2× 1mol H2/2.016g H2 × 3mol H2SO4/2molH2 = 7.44molH2SO4

You could go further in the analysis, but you are given the molarity of the sulfuric acid. That should give you some guidance in the direction to solve this problem.

Therefore, using the M=mol/L equation we can plug in the number and solve for the volume.

6.0L=7.44mol/L = (7.44mol H2SO4)6.0MH2SO4.

=44.64LH2SO4