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12 January, 16:28

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.

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  1. 12 January, 16:54
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    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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