Đáp án:
`C%_(ZnCl_2)~~6,47%`
Giải thích các bước giải:
`Zn+2HCl->ZnCl_2+H_2`
`n_(Zn)=m/M=26/65=0,4(mol)`
`m_(HCl)=(C%.mdd)/(100%)=(3,65%.500)/(100%)=18,25(g)`
`=>n_(HCl)=m/M=(18,25)/(1+35,5)=0,5(mol)`
So sánh tỉ lệ`: (n_(Zn))/1=(0,4)/1>(n_(HCl))/2=(0,5)/2`
`=>Zn` dư`, HCl` hết
Theo `PTHH: n_(ZnCl_2)=1/2n_(HCl)=1/(2).0,5=0,25(mol)`
`=>m_(ZnCl_2)=n.M=0,25(65+35,5.2)=34(g)`
Theo `PTHH: n_(H_2)=1/2n_(HCl)=1/(2).0,5=0,25(mol)`
`=>m_(H_2)=n.M=0,25.1.2=0,5(g)`
`=>mdd_(ZnCl_2)=m_(Zn)+mdd_(HCl)-m_(H_2)=26+500-0,5=525,5(g)`
`=>C%_(ZnCl_2)=(mct)/(mdd).100%=34/(525,5).100%~~6,47%`
