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25 October, 19:44

If 2 CH3OH + 3 O2 - > 2CO2 + 4 H2O was carried out in the laboratory and 219 g of water was produced, what would the percent yield be if the reaction started with 229 g of methanol in excess oxygen gas

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  1. 25 October, 23:08
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    Percent yield = 84.5 %

    Explanation:

    Given dа ta:

    Mass of methanol = 229 g

    Actual yield of water = 219 g

    Percent yield of water = ?

    Solution:

    Chemical equation:

    2CH₃OH + 3O₂ → 2CO₂ + 4H₂O

    Number of moles of methanol:

    Number of moles = mass / molar mass

    Number of moles = 229 g / 32 g/mol

    Number of moles = 7.2 mol

    Now we will compare the moles of water with methanol.

    CH₃OH : H₂O

    2 : 4

    7.2 : 4/2*7.2 = 14.4 mol

    Mass of water:

    Mass = number of moles * molae mass

    Mass = 14.4 mol * 18 g/mol

    Mass = 259.2 g

    Percent yield:

    Percent yield = actual yield / theoretical yield * 100

    Percent yield = 219 g / 259.2 g * 100

    Percent yield = 84.5 %
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