\(H_2+Cl_2\rightarrow2HCl\)
\(AgNO_3+HCl\rightarrow AgCl+HNO_3\)
\(n_{H2}=\frac{0,672}{22,4}=0,03\left(mol\right)\)
\(n_{CL2}=\frac{0,56}{22,4}=0,025\left(mol\right)\)
\(\rightarrow\) Tính theo Cl2
\(n_{AgCl}=\frac{1,435}{143,5}=0,01\left(mol\right)\)
\(\rightarrow n_{HCl_{bđ}}=0,02\left(mol\right)\)
\(\rightarrow n_{HCl_{lt}}=0,025.2=0,05\left(mol\right)\)
\(\rightarrow H=\frac{0,02}{0,05}.100=40\%\)