\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{109,5.20\%}{36,5}=0,6\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,2}{1}< \dfrac{0,6}{2}\Rightarrow HCldư\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ n_{HCl\left(dư\right)}=0,6-0,2.2=0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,m_{ZnCl_2}=136.0,2=27,2\left(g\right)\\ c,m_{ddsau}=13+109,5-0,2.2=122,1\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{27,2}{122,1}.100\approx22,277\%\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,2.36,5}{122,1}.100\approx5,979\%\)
Zn + 2HCl -> ZnCl2 + H2
a, nZn = 13/65= 0,2(mol)
mHCl= 109,5.20%/100%=21.9(g)
nHCl=21,9/36,5=0,6(mol)
Theo PT nHCl = 2nZn= 2.0,2= 0,4(mol)<0,6(mol)
=> HCl pư dư, Zn pư hết
Theo PT: nH2= nZn =0,2(mol)
VH2=0,2.22,4=4,48(l)
b, Theo PT: nZnCl2=nZn=0,2(mol)
mZnCl2= 0,2.136=27,2(g)
c, mdd sau pư= 13+109,5-0,2.2=122,1(g)
C%dd ZnCl2=27,2.100%/122,1=22,28%
nHCl dư= 0,6-0,4=0,2(mol)
mHcl
