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.

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