`PTHH:Zn+2HCl→ZnCl_2+H_2↑`
a)Ta có: `n_(Zn)= (6,5)/65= 0,1mol`
Theo `PTHH`,ta thấy: `n_(HCl)=2.n_(Zn)=2.0,1=0,2(mol)`
`⇒mHCl = 0,2.36,5= 7,3(g)`
`⇒C%HCl=100%.(7,3)/50=14,6%`
b)Theo `PTHH`,ta thấy: `n_(H_2)=n_(Zn)=n_(ZnCl_2)=0,1(mol)`
`⇒m_(H_2)=n.M=0,1.2=0,2(g)`
Theo Định luật bảo toàn khối lượng,ta có:
` m_(Zn)+m_(ddHCl)=m_(ddZnCl_2)+m_(H_2)`
`⇒m_(ddZnCl_2)=(6,5+50)-0,2=56,3(g)`
`⇒m_(ZnCl_2)=n.M=0,1.136=13,6(g)`
`⇒C%ZnCl2=100%.(13,6)/(56,3)≈24,16%`
Xin hay nhất =_=