\(H_2+Cl_2\left(0,18\right)\rightarrow2HCl\left(0,36\right)\)
\(n_{H_2}=\frac{10}{22,4}\approx0,446\left(mol\right)\)
\(n_{Cl_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
Vì \(\frac{n_{H_2}}{1}=0,446>\frac{n_{Cl_2}}{1}=0,3\) nên ta tính HCl theo Cl2
Số mol Cl2 tham gia phản ứng là: \(0,3.60\%=0,18\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,36.36,5=13,14\left(g\right)\)
\(\Rightarrow m_{HCl\left(conlai\right)}=13,14.\left(100\%-5\%\right)=12,483\left(g\right)\)