Question

20g含锌质量分数为65%的锌片（杂质不与酸反应）能与50g稀硫酸恰好完全反应.计算参加反应的稀硫酸中硫酸

Answer 1

m(Zn)=a(Zn)m(杂Zn)=65%*20g=65*0.2g

n(Zn)=m(Zn)/M(Zn)=65*0.2g/65g/mol=0.2mol

由Zn+H2SO4=H2+ZnSO4

n(H2SO4)=0.2mol

m(H2SO4)=0.2mol*98g/mol=19.6g