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13 December, 10:44

The mass of h2 produced by reaction of 1.80 g al and 6.00 g h2so4 is 0.112 g. what is the percent yield?

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  1. 13 December, 14:40
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    First, let us write the reaction. We know that the reactants are Aluminum and Sulfuric acid, and one of the products is hydrogen. That means the reaction is a single replacement reaction as shown below:

    2 Al + 3 H₂SO₄ = 3 H₂ + Al₂ (SO₄) ₃

    Next, let us determine which of the reactants is limiting:

    1.80 g Al (1 mol/26.98 g) (3 mol H₂SO₄/2 mol Al) (98 g H₂SO₄/mol) = 9.807 g

    It would need 9.807 g. But the only available amount is 6 g. That means that H₂SO₄ is the limiting reactant. We use 6 g as the basis to know the theoretical yield:

    6 g * (1 mol H₂SO₄/98 g) * (3 mol H₂ / 3 mol H₂SO₄) * (2 g / mol H₂) = 0.122 g

    Therefore, the percent yield is equal to

    Percent yield = (0.112 g/0.122 g) * 100

    Percent yield = 91.8%
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