1.80 grams of an impure mixture containing sodium carbonate required 84.0 mL of 0.125 M H₂SO₄(aq)for complete neutralization.What percent of the mixture is sodium carbonate? A)62% B)31% C)120% D)57% E)1.1%

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Quizplus · Accepted Answer

The reaction between sodium carbonate (Na2CO3) and sulfuric acid (H2SO4) is: Na2CO3 + H2SO4 → Na2SO4 + CO2 + H2O. The moles of H2SO4 used are 0.084 L * 0.125 mol/L = 0.0105 mol. Since the reaction ratio is 1:1, moles of Na2CO3 = 0.0105 mol. The mass of Na2CO3 is 0.0105 mol * 106 g/mol = 1.113 g. The percent of Na2CO3 in the mixture is (1.113 g / 1.80 g) * 100% = 61.83%, approximately 62%.

180 Grams of an Impure Mixture Containing Sodium Carbonate Required