Question

. A mixture containing 50.0 g of hydrogen gas and 50.0 g of oxygen gas is sparked and allowed to react.
Which reactant is the limiting reactant, how much water is produced, and how much of the excess
reactant remains?

Answer 1

m(O2)=50g
m(H2)=50g
m(H2O)-?
n2(H2)-?
2H2 + O2 = 2H2O
n(O2)= m (O2)/M(O2) =50g / 32 g/mol= 1,56 mol.
n(H2)= m (H2)/M(H2) =50g / 2 g/mol= 25 mol.
The oxygen gas is the limiting reactant, because n(O2)< 2 n(H2) from reaction.
n(H2O)= 2n(O2)= 2n (H2 reac.)=2*1,56 mol=3,12mol.
m(H2O)=n(H2O)*M(H2O)= 3,12mol* 18 g/mol.
n2(H2)= n(H2) - n (H2 reac.)=25mol - 3,12mol=21,88mol.