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13 January, 18:03

What mass of MgCl2 was produced from 5.87 of Mg (OH) 2 reacting with 12.84 g of HCl? Identify the limiting and the excess reactants.

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  1. 13 January, 18:18
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    Answers;

    Mass of MgCl2 = 9.615 g

    Mg (OH) ₂ i limiting reactant

    HCl is the excess reactant

    Explanation;

    The equation for the reaction is;

    Mg (OH) 2 + 2 HCl → MgCl2 + H2O

    5.87 g Mg (OH) 2x (1mol Mg (OH) 2/58.31g) = 0.101 moles OF Mg (OH) 2

    Moles of MgCl2 = 0.101 moles

    Mass of MgCl2 = 0.101 * 95.2g/mole

    = 9.615 g MgCl2

    Moles of HCl

    0.101 x (2 mol HCl / 1 mol Mg (OH) 2) = 0.202 moles

    Mass of HCl

    0.202 moles x (36.45 g HCl/1 mol HCl) = 7.34 g

    Therefore; 7.34 g of HCl are needed out of 12.84 g HCl given,

    Therefore; Mg (OH) 2 is the limiting reactant and HCl is the excess reactant
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