Question

5.6g的铁粉与100g稀硫酸恰好完全反应,求

Answer 1

Fe + H2SO4 == FeSO4 + H2↑

56 98 152 2

5.6g m1 m2 m3

56/5.6 = 98/m1 = 152/m2 = 2/m3

得m1 = 9.8g,m2 = 15.2g,m3 = 0.2g

所以

该稀硫酸的溶质质量分数 = 9.8/100 = 9.8%

反应后所得溶液质量 = 5.6+100-0.2 = 105.4g

反应后所得溶液的溶质质量分数 = 15.2/105.4 = 14.4%