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5 February, 19:03

In the reaction shown above, 2.00 x 10^3 g caco3 produce 1.05 x 10^3 g of cao, what is the percent yield?

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  1. 5 February, 19:30
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    First, we have to get the theoretical yield of CaO:

    the balanced equation for the reaction is:

    CaCO3 (s) →CaO (s) + CO2 (g)

    covert mass to moles:

    moles CaCO3 = mass of CaCO3 / molar mass of CaCO3

    = 2x10^3 / 100 = 20 moles

    the molar ratio between CaCO3 : CaO = 1:1

    ∴moles of CaO = 1 * 20 = 20 moles

    ∴mass of CaO = moles of CaO * molar mass of CaO

    = 20 * 56 = 1120 g

    ∴the theoritical yield = 1120 g and we have the actual yield = 1.05X10^3

    ∴Percent yield = actual yield / theoritical yield * 100

    = (1.05x10^3) / 1120 * 100

    = 94 %
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