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12 December, 06:09

Calculate the mass of water produced when 1.92 g of butane reacts with excess oxygen.

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  1. 12 December, 08:22
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    C4h10+6.5o2=4co2+5h2o

    moles of butane=1.92/58=0.0331 moles

    moles of water=0.1655 moles/

    as the butane and water has 1 is to 5 molar ratio

    0.1655=mass/18

    mass=2.98 g

    mass of water produced = 2.98 g
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