\(m_{HCl}=\dfrac{100\times7,3\%}{100\%}=7,3\left(g\right)\)
\(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Mol (pt): 1 2 1 1
Mol (đề): 0,2
Theo pt, ta có: \(n_{ZnCl_2}=n_{Zn}=n_{H_2}=\dfrac{n_{HCl}}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,1\cdot65=6,5\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\\m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\end{matrix}\right.\)
Theo ĐLBTKL, ta có: \(m_{\text{dd }ZnCl_2}=m_{Zn}+m_{\text{dd }HCl}-m_{H_2}\)
\(=6,5+100-0,2=106,3\left(g\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{13,6}{106,3}\cdot100\%\approx12,79\%\)
