2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)
1) 2 reactants are given, so we have to find which of them is an excess reactant and which of them will react completely.
Convert gram of Na and H2O into moles.
M(Na) =23.0 g/mol
10.0 g Na*1 mol Na/23.0 g Na≈0.4349 mol Na
M(H2O) = 2*1.0 +16.0 = 18.0 g/mol
50.0 g H2O*1 mol H2O/18.0 g H2O =2.778 mol H2O
2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)
from equation 2 mol 2 mol
given 0.4349 mol 2.778 mol
We can see that H2O is an excess reactant, so we are going to find amount of H2 using Na.
2)
2Na(s) + 2 H2O (liquid) -> 2 NaOH (aq) + H2 (g)
from equation 2 mol 1 mol
given 0.4349 mol x mol
x=(0.4349*1)/2 =0.21745 mol H2
3) mass of H2
0.21745 mol H2 * 2.0 g H2/1 mol ≈ 0.435 g H2
Answer : 0.435 g H2.
