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24 September, 14:09

If 638.44g CuSO4 reacts with 240.0g NaOH, which is the limiting reactant?

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  1. 24 September, 15:04
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    CuSO₄ + 2NaOH → Cu (OH) ₂ + Na₂SO₄

    M (CuSO₄) = 159.61 g/mol

    m (CuSO₄) = 638.44 g

    n (CuSO₄) = m (CuSO₄) / M (CuSO₄)

    n (CuSO₄) = 638.44/159.61 = 4.00 mol

    M (NaOH) = 40.00 g/mol

    m (NaOH) = 240.0 g

    n (NaOH) = m (NaOH) / M (NaOH)

    n (NaOH) = 240.0/40.00 = 6.00 mol

    on the reaction equation

    CuSO₄ : NaOH = 1 : 2

    in practice

    CuSO₄ : NaOH = 4 : 6 = 1 : 1.5 ⇒ NaOH is the limiting reactant
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