Question

In the reaction Mg (s) + 2HCl (aq) -> H2 (g) + MgCl2 (aq), how many grams of hydrogen gas will be produced from 300 milliliters of a 4 M HCl in an excess of Mg? Please show all work

Answer 1

Mg(s) + 2HCl(aq) = H₂(g) + MgCl₂(aq)

v=300 ml=0.3 l
c=4 mol/l

n(HCl)=vc

m(H₂)/M(H₂)=n(HCl)/2

m(H₂)=M(H₂)vc/2

m(H₂)=2.0g/mol·0.3 l · 4mol/l /2 = 1.2 g

1.2 grams of hydrogen gas will be produced