设FeSO4质量是x,H2质量是yFe + H2SO4 = FeSO4 + H2↑56 152 25.6 x y56/152= 5.6/x 解得x= 15.2 g56/2= 5.6/y 解得x= 0.2 g溶质质量 = 15.2 g溶质质量分数 = 15.2 / (100+5.6-0.2) * 100% = 14.4%...