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30 August, 11:14

How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate is combined with 15.0 g of hydrochloric acid?

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  1. 30 August, 12:03
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    22.7 g of CaCl₂ are produced in the reaction

    Explanation:

    This is the reaction:

    CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

    Now, let's determine the limiting reactant.

    Let's divide the mass between the molar mass, to find out moles of each reactant.

    29 g / 100.08 g/m = 0.289 of carbonate

    15 g / 36.45 g/m = 0.411 of acid

    1 mol of carbonate must react with 2 moles of acid

    0.289 moles of carbonate will react with the double of moles (0.578)

    I only have 0.411 of HCl, so the acid is the limiting reactant.

    Ratio is 2:1, so I will produce the half of moles, of salt.

    0.411 / 2 = 0.205 moles of CaCl₂

    Mol. molar mass = mass → 0.205 m. 110.98 g/m = 22.7 g
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