What mass of calcium carbonate in grams will be produced from 1.67 mole of sodium carbonate
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Expert's answer
2015-10-23T12:02:24-0400
Ca(OH)2(aq) + Na2CO3 (aq) = CaCO3(s) + 2NaOH(aq) From the stoichiometry of this reaction it is obviously that 1 mole of sodium carbonate yields to 1 mole of calcium carbonate, so n(CaCO3) = n(Na2CO3) = 1.67 mol The relationship between the amount of substance and the mass is as follows: m(CaCO3) = n(CaCO3) M(CaCO3) m(CaCO3) = 1.67x100.09 = 167.15 g
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