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12 November, 15:11

What mass of o2 is required to produce 14.5 g of co2 if the reaction has a 65.0% yield? Ch4 (g) + 2o2 (g) / longrightarrow ⟶ co2 (g) + 2h2o (g) ?

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  1. 12 November, 15:23
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    the balanced equation for the combustion of methane is as follows

    CH₄ + 2O₂ - -> CO₂ + 2H₂O

    stoichiometry of O₂ to CO₂ is 2:1

    the percentage yield of CO₂ is 65.0 %

    percentage yield = (actual yield / theoretical yield) x 100 %

    substituting these values to find the theoretical yield

    65 % = (14.5 g / theoretical yield) x 100 %

    theoretical yield = 14.5 / 0.65 = 22.3 g

    the theoretical yield of CO₂ is 22.3 g

    therefore number of moles of CO₂ that was expected to be formed - 22.3 g / 44 g/mol = 0.507 mol

    according to molar ratio O₂ to CO₂ ratio is 2:1

    therefore number of O₂ moles required to react - 0.507 x2 = 1.01 mol

    mass of O₂ required - 1.01 mol x 32 g/mol = 32.32 g

    therefore mass of O₂ needed is 32.32 g
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