What mass of Al is necessary?

Question

What mass of Al is necessary to produce 1.32 L of H2 at STP in the following reaction? 2 Al + 3 H2SO4 ® Al2(SO4)3 + 3 H2

Answer 1

1.32LH2 x 1molH2/22.4LH2 x 2molAl/3molH2 x 27gAl/1molAl = (1.32)(2)(27)/(22.4)(3) = 1 g

The 1.32L H2 is given. The first factor comes from the molar volume at STP. The L H2 cancel, leaving moles H2. The next factor comes from the balanced equation. The moles H2 cancel, leaving moles Al. The last factor comes from the atomic weight of Al=27. The moles Al cancel, leaving g Al, which was desired.