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11 July, 09:12

How many grams of mgcl2 will be produced from 10.5g of mg (oh) 2 and 41g of hcl?

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  1. 11 July, 09:18
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    The number of grams of MgCl2 that will be produced from 10.5 g Mg (OH) 2 and 41 of HCl is 17.154 grams

    calculation

    write the equation for reaction

    Mg (OH) 2 + HCl → MgCl2 + 2H2O

    find the moles of each compound used

    moles=mass / molar mass

    moles of Mg (OH) 2 = 10.5g / 58.3 g/mol = 0.180 moles

    moles of HCl = 41 g / 36.5g/mol = 1.123 moles

    by use of of mole ratio between Mg (OH) 2 to MgCl2 which is 1:1 moles of MgCl2 is also 0.180 moles

    by use of mole ratio between HCl to MgCl2 which is 2:1 the moles of MgCl2 = 1.123 x1/2 = 0.5615 moles

    HCl was in excess while Mg (OH) 2 was the limiting reagent, therefore moles of MgCl2 = 0.180 moles

    mass of MgCl2 = moles MgCl2 x molar mass of MgCl2 (24.3 + 35.5 x2=95.3)

    mass = 0.180 x 95.3 = 17.154 grams
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