某纯碱样品中含有杂质NaCl，为测定该样品中Na2CO3的质量分数，称取纯碱样品6 g放入20 g水中使其完全溶解，再加入稀盐酸26.2 g，恰好完全反应，反应后溶液的总质量为50 g。
试计算：(写出计算过程，结果保留一位小数)
(1)生成二氧化碳的质量;
(2)样品中Na2CO3的质量分数;
(3)反应后所得溶液中溶质的质量分数。
解：(1)根据质量守恒定律，生成CO2的质量为20 g+6 g+26.2 g-50 g=2.2 g。 (2)设参加反应的Na2CO3的质量为x，生成氯化钠的质量为y。
Na2CO3+2HCl===2NaCl+CO2↑+H2O
106 117 44
x y 2.2 g
10644=x2.2 g x=5.3 g
11744=y2.2 g y=5.85 g
则样品中Na2CO3的质量分数为5.3 g6 g×100%≈88.3%。
(3)反应后溶液中溶质NaCl的质量为5.85 g+(6 g-5.3 g)=6.55 g。
则反应后所得溶液中溶质的质量分数为6.55 g50 g×100%=13.1%。
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