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Chemistry

2Ca(s) + O2(g) → 2CaO(s)

Calculate the mass of calcium oxide that can be prepared from 7.52 g of Ca and 5.01 g of O2.

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  1. 7.52g of Ca = 0.188 moles
    5.01g of CO2 = 0.114 moles
    The equation says that 0.188 moles of Ca will use 0.094 moles of CO2, and produce 0.188 moles of CaO = 10.54g

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  2. oops!! oobleck misread the problem as CO2 instead of O2.
    mols Ca = 7.52/40 = 0.188
    mols O2 = 5.01/32 = 0.156
    The problem tells you that 0.188 g Ca will use 0.094 mols O2 and produce 0.188 moles of CaO
    0.188 mols CaO x 56 g CaO/mol CaO = 10.5 g CaO so the answer comes out the same although the moles O2 are not the same as moles CO2. Some people are just born lucky. BTW, the limiting reagent is Ca and the excess reagent is O2.

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