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20 April, 13:18

In a reaction between 6.0 g of oxygen gas, 4.0 g of hydrogen gas, and 5.0 g of solid sulfur at standard temperature and pressure to make H₂SO₄, which is the limiting reagent?

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  1. 20 April, 15:10
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    The limiting reagent is the O₂

    Explanation:

    We can think, this reaction

    2O₂ (g) + H₂ (g) + S (s) → H₂SO₄

    Mole of each = Mass / molar mass

    6 g / 32 g/m = 0.187 mole O₂

    4g / 2 g/m = 2 mole H₂

    5g / 32.06 g/m = 0.156 mole S

    Ratio between reactants is 2:1:1, 1:2:1, 1:1:2

    For 2 mole of O₂, I need to react 1 mol of H₂ and 1 mol of S

    0.187 mole of O₂, I need (the half)

    0.093 mole of H₂ and 0.093 mole of S

    For 1 mole of H₂, I need to react 2 mole of O₂ and 1 mol of S

    2 mole of H₂, I need (the double of O₂ and the same for S)

    4 mole of O₂; 2 mole of S

    For 1 mol of S, I need to react 1 mol of H₂ and 2 mole of O₂

    0.156 mole I need the same amount for H₂ and the double for O₂

    0.156 mole of H₂ and 0.312 mole of O₂

    In both cases, I can't make react, all the mass of oxygen, so this is the limiting reagent.
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