Sodium and water react according to the equation

2Na (s) + 2 H2O (liquid) - > 2 NaOH (aq) + H2 (g).

If 10.0 g id sodium is placed in 50.0 g of water, calculate the mass of hydrogen that will be produced.

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.