Bài giải :
Câu 7 :
a.
`-n_{Zn}=\frac{119,5}{65}≈1,838(mol)`
`Zn+2HCl→ZnCl_2+H_2↑`
1,838 → 3,676 1,838 1,838 (mol)
b.
`-V_{H_2}(đktc)=1,838.2,4=41,1712(l)`
c.
$-m_{chất..tan..HCl}=3,676.36,5=134,174(g)$
`⇒m_{dd..HCl}=\frac{134,174.100%}{10%}=1341,74(g)`
d.
- Dung dịch thu được sau phản ứng: `ZnCl_2`
$-m_{dd..sau..pứ}=m_{Zn}+m_{dd..HCl}-m_{H_2}$
$=119,5+1341,74-1,838.2=1457,564(g)$
`-m_{chất..tan..ZnCl_2}=1,838.136=249,968(g)`
`⇒C%ZnCl_2=\frac{249,968}{1457,564}.100%≈17,15%`
