Zn+H2SO4=ZnSO4+H2 65 98 161 2 x 100g*9.8% y zx=6.5g; y=16.1g ; z=0.2g反应后所得的溶液中溶质的质量分数是：16.1g/(6.5g+100g-0.2g)*100%=15.146%
