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19 August, 20:34

When 412.5g of calcium carbonate react with 521.9g of aluminum fluoride, how many grams of each product can be produced

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  1. 19 August, 20:41
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    Answer;

    321.8 g CaF2

    321.5 g Al2 (CO3) 3

    Explanation;

    The equation for the reaction is;

    3 CaCO3 + 2 AlF3 → 3 CaF2 + Al2 (CO3) 3

    Number of moles of CaCO3 will be;

    = (412.5 g CaCO3) / (100.0875 g CaCO3/mol)

    = 4.12139 mol CaCO3

    Number of moles of AlF3 will be;

    = (521.9 g AlF3) / (83.9767 g AlF3/mol)

    = 6.21482 mol AlF3

    But;

    4.12139 moles of CaCO3 would react completely with 4.12139 x (2/3) = 2.74759 moles of AlF3.

    Thus; there is more AlF3 present than that, so AlF3 is in excess, and CaCO3 is the limiting reactant.

    Therefore;

    Mass of CaF2 will be;

    (4.12139 mol CaCO3) x (3/3) x (78.0752 g CaF2/mol) = 321.8 g CaF2

    Mass of Al2 (CO3) 3 on the other hand will be;

    (4.12139 mol CaCO3) x (1/3) x (233.9903 g Al2 (CO3) 3/mol) = 321.5 g Al2 (CO3) 3
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