Đáp án+Giải thích các bước giải:
Zn+2HCl→ZnCl_{2}+H_{2}
n_{Zn}=13/65=0,2(mol)
n_{HCl}=0,2.1=0,2(mol)
Ta thấy : n_{Zn}>n_{HCl}/2
→Zn dư 
a.
n_{ZnCl_{2}}=n_{H_{2}}=1/2.n_{HCl}=(1)/(2).0,2=0,1(mol)
→C_{M}ZnCl_{2}=(0,1)/(0,2)=0,5(M)
V_{H_{2}}=x=0,1.24,79=2,479(l)
b.
m_{dd spứ}=0,1.65+200.1,1-0,1.2=226,3(g)
m_{ZnCl_{2}}=0,1.136=13,6(g)
C%_{ZnCl_{2}}=(13,6)/(226,3).100%=6%

a)
n_{Zn}={13}/{65}=0,2(mol)
n_{HCl}=0,2.1=0,2(mol)
Zn+2HCl->ZnCl_2+H_2
0,1←0,2->0,1->0,1(mol)
0,2>{0,2}/2->Zn dư
C_{M\ ZnCl_2}={0,1}/{0,2}=0,5M
x=V_{H_2}=0,1.24,79=2,479(L)
b)
m_{dd\ sau}=0,1.65+200.1,1-0,1.2=226,3(g)
C\%_{ZnCl_2}={0,1.136}/{226,3}.100\%=6\%