Mg metal reacts with HCl. If 1.320 of Mg is added to 50mL of a 3M HCl solution, how many g of H2 form?

Answer 1

Mg + 2HCl ===> MgCl2 + H2

Atomic weights: Mg=24 Cl=35.5 H=1

Let the HCl solution be called S

1.320gMg x 1molMg/24gMg = 0.055 mol Mg

50mLS x 3molHCl/1000mLS = 0.15 mol HCl

According to the balanced equation, 0.055 mol Mg would require 0.110 mol HCl. You have more HCl than that, so Mg is the limiting reagent. Going back to the moles Mg above:

0.055molMg x 1molH2/1molMg x 2gH2/1molH2 = 0.11g H2