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7 September, 22:02

Consider the reaction of CaC2 and water to produce CaCO3 and NH3 according to the reaction CaCN2 + 3H2O → CaCO3 + 2 NH3. How much CaCO3 is produced upon reaction of 45 g CaCN2 and 45 g of H2O

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  1. 7 September, 23:06
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    16.27 g of CaCO3 are produced upon reaction of 45 g CaCN2 and 45 g of H2O.

    Explanation:

    Ca (CN) 2 + 3H2O → CaCO3 + 2 NH3

    First of all, let's find out the limiting reactant.

    Molar mass Ca (CN) 2.

    Molar mass H2O: 18 g/m

    Moles of Ca (CN) 2: mass / molar mass

    45 g / 92.08 g/m = 0.488 moles

    Moles of H2O: mass / molar mass

    45g / 18g/m = 2.50 moles

    This is my rule of three

    1 mol of Ca (CN) 2 needs 3 moles of H2O

    2.5 moles of Ca (CN) 2 needs (2.5. 3) / 1 = 7.5 moles

    I need 7.5 moles of water, but I only have 0.488. Obviously water is the limiting reactant; now we can work on it.

    3 moles of water __ makes __ 1 mol of CaCO3

    0.488 moles of water __ makes ___ (0.488. 1) / 3 = 0.163 moles

    Molar mass CaCO3 = 100.08 g/m

    Molar mass. moles = mass

    100.08 g/m. 0.163 moles = 16.27 g
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