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17 March, 18:32

Calcium reacts with sulfur forming calcium sulfide. What is the theoretical yield (g) of CaS (s) that could be prepared from 8.54 g of Ca (s) and 2.33 g of sulfur (s) ? Do not type units with your answer.

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  1. 17 March, 22:14
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    5.242

    Explanation:

    Molar mass of Ca = 40.078 g/mol

    Molar mass of S = 32.065 g/mol

    Equation of reaction

    Ca (s) + S (s) - CaS (s)

    To find the limiting reagent

    Mole of Ca = 8.54/40.078 (g/mol) = 0.213 mol

    Mole of S = 2.33g / 32.065 (g/mol) = 0.072664 mol

    From their mole ratio, sulphur is the limiting reagent

    From the reaction

    1 mole of calcium react with 1 mole of Sulphur to yield 1 mole of calcium sulphide

    0.072664 mol of calcium will react with 0.072664 of sulphur to yield 0.072664 mole of CaS

    Theoretical yield of CaS = 0.072664 * molar mass of CaS = 0.072664 * 72.143 g/mol = 5.242 g
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