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Percentage yield calculation with gases

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    • Thread Starter

    Hello, I'm currently revising unit 1 as a resit. Been fine with all the other multiple choice questions but this one (from last year's paper):
    Lithium reacts with water to produce hydrogen.Li(s) + H2O(l) → LiOH(aq) + ½H2(g)

    (a) In an experiment, 0.069 g (0.01 mol) of lithium produced 90 cm3 of hydrogen at room temperature and pressure. What is the percentage yield of hydrogen?

    [1 mol of any gas occupies 24 dm3 at room temperature and pressure.]
    I can't seem to get the right answer, which was 75%. I managed to get 75% after looking at the answer, but the method I used is unusual...

    Is there an actual method for doing this, without me doing these random calculations that somehow got the answer (which I don't understand how)?

    I've done this weird method which involved me doing
    0.01 * 24 = 0.24
    0.24 * 1000 = 240
    90/240 = 0.375 * 2 = 0.75
    0.75*100 = 75% which is the right answer

    Ratio of Lithium to Hydrogen gas is 1:0.5

    Therefore, the moles of Hydrogen gas is half the moles of lithium. ( moles = 0.005)

    volume of gas = moles * 24000 = 0.005 * 24000 = 120cm3

    percentage yield = (90/120) * 100 = 75%
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