根据质量守恒,反应生成的BaSO4沉淀质量 = 200+100-276.7 = 23.3g
Na2SO4+BaCl2==BaSO4↓+2NaCl
142 233 117
x 23.3g y
142/x = 233/23.3 = 117/y
x = 14.2g,y = 11.7g
所以原硫酸钠溶液中溶质的质量分数 = 14.2/200 = 7.1%
滤液中的溶质的质量分数 = 11.7/276.7 = 4.2%
根据质量守恒,反应生成的BaSO4沉淀质量 = 200+100-276.7 = 23.3g
Na2SO4+BaCl2==BaSO4↓+2NaCl
142 233 117
x 23.3g y
142/x = 233/23.3 = 117/y
x = 14.2g,y = 11.7g
所以原硫酸钠溶液中溶质的质量分数 = 14.2/200 = 7.1%
滤液中的溶质的质量分数 = 11.7/276.7 = 4.2%