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Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid water H2O. Suppose 59. g of sulfuric acid is mixed with 86.9 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits.

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jbferrado

Answer:

22 g of water

Explanation:

The balanced chemical equation is:

H₂SO₄(aq) + 2 NaOH(aq) → Na₂SO₄(aq) +2 H₂O(l)

⇒ 1 mol of H₂SO₄ reacts with 2 moles of NaOH to produce 2 moles of H₂O.

First, we have to convert the moles to grams by using the molecular weight (Mw) of each compound:

  • Mw H₂SO₄ = (1 g/mol x 2 H) + 32 g/mol S + (16 g/mol x 4 O) = 98 g/mol

        mass of H₂SO₄ = moles x Mw = 1 mol x 98 g/mol = 98 g

  • Mw NaOH = 23 g/mol Na + 16 g/mol O + (1 g/mol H) = 40 g/moL

        mass of NaOH = 2 moles x 40 g/mol = 80 g

  • Mw H₂O = (1 g/mol x 2 H) + 16 g/mol O = 18 g/moL

        mass of NaOH = 2 moles x 18 g/mol = 36 g

Now, we have to determine which is the limiting reactant. The stroichiometric ratio is 98 g H₂SO₄/80 g NaOH. We have 86.9 g of NaOH, so we need the following amount of H₂SO₄:

98 g H₂SO₄/80 g NaOH x 86.9 g NaOH = 106.4 g H₂SO₄

As we need more H₂SO₄ (106.4 g) than we have (59 g), the limiting reactant is H₂SO₄.

Finally, we calculate the mass of H₂O produced with the limiting reactant. The stoichiometric ratio is 36 g H₂O/98 g H₂SO₄, so we multiply the actual mass of H₂SO₄ by the stoichiometric ratio to calculate the maximum mass of H₂O produced:

mass of H₂O = 36 g H₂O/98 g H₂SO₄ x 59 g H₂SO₄ = 21.7 g ≅ 22 g

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