a)
`PTHH:Zn+2HCl->ZnCl_2+H_2↑`
b)
Ta có: `n_(Zn)=m/M=(1,2)/65≈0,02(mol)`
Theo `PTHH`,ta thấy: `n_(Zn)=n_(H_2)=0,02(mol)`
`=>V_(H_2)=n.22,4=0,02.22,4=0,448(l)`
c)
Theo `PTHH`,ta thấy: `n_(HCl)=2.n_(Zn)=2.0,02=0,04(mol)`
`=>m_(HCl)=n.M=0,04.36,5=1,46(g)`
`->C%ddHCl=(1,46)/200×100%=0,73%`
d)
Theo Định luật bảo toàn khối lượng:
`m_(Zn)+m_(ddHCl)=m_(ddZnCl_2)+m_(H_2)`
`=>m_(ddZnCl_2)=(m_(Zn)+m_(ddHCl))-m_(H_2)`
`=>m_(ddZnCl_2)=(1,2+200)-(0,02.2)`
`=>m_(ddZnCl_2)=201,16`
Theo `PTHH`,ta thấy: `n_(ZnCl_2)=n_(Zn)=0,02(mol)`
`=>m_(ZnCl_2)=n.M=0,02.136=2,72(g)`
`->C%ddZnCl_2=(2,72)/(201,16)×100%≈1,4%`