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2 March, 00:53

The combustion of 0.136 kg of methane in the presence of excess oxygen produced 353 g of carbon dioxide what is the percentage yield

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  1. 2 March, 01:08
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    The combustion reaction for methane is as follows

    CH4 + 2O2 - > CO2 + 2H2O

    Stoichiometry of methane to CO2 is 1:1

    Since oxygen is in excess, methane is the limiting reactant therefore amount of CO2 produced depends on methane reacted,

    Number of methane moles reacted = 136 g / 16 g/mol = 8.5 mol

    Therefore moles of CO2 to be produced = 8.5 mol

    Theoretically mass of CO2 to be produced = 8.5 mol x 44 g/mol = 374 g

    However actual yield = 353 g

    Percent yield = actual yield / theoretical yield x 100%

    Percent yield = 353 g / 374 g x100%

    Percentage yield of CO2 = 94.4%
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