\(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)
\(n_{HCl}=\frac{150.3,65}{100.36,5}=0,15\left(mol\right)\)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
TheoPT: 1mol.....2mol
TheoĐB:0,1mol...0,15mol
Lập tỉ lệ: \(\frac{0,1}{1}>\frac{0,15}{2}\)
=>Zn dư, HCl phản ứng hết
Theo PT: \(n_{Zn\left(pu\right)}=\frac{1}{2}n_{HCl}=0,075\left(mol\right)\)
\(\Rightarrow n_{Zn\left(dư\right)}=0,1-0,075=0,025\left(mol\right)\)
\(\Rightarrow m_{Zn\left(dư\right)}=0,025.65=1,625\left(g\right)\)
b) Sau p/ứ dd là ZnCl2
\(m_{ddZnCl_2}=m_{Zn}+m_{ddHCl}-m_{Zn\left(dư\right)}-m_{H_2}\)
\(\Leftrightarrow m_{ddZnCl_2}=6,5+150-1,625-0,075.2\)
\(\Leftrightarrow m_{ddZnCl_2}=154,725\left(g\right)\)
Theo PT: \(n_{ZnCl_2}=\frac{1}{2}n_{HCl}=0,075\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,075.136=10,2\left(g\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\frac{10,2}{154,725}.100\approx6,592\%\)
