化学计算一道 快某化学样品由氯化钠和碳酸钠组成.取该样品6克放于烧杯中,用44克水完全溶解,再想烧杯中滴加稀盐酸至不再有

m(CO2) = 6+44+20-67.8 = 2.2g
Na2CO3 +2HCl = 2NaCl + CO2 + H2O
106.117.44
x.y.2.2
106:x = 117:y = 44:2.2
x = 5.3g
y = 5.85g
m(NaCl) = (6-5.3)+5.85 = 6.55g
1.原样品中碳酸钠的质量分数=100%*5.3/6 = 88.3%
2.反应后所得溶液中溶质的质量分数 = 100%*6.55/67.8 = 9.66%