\(PTHH:H_2+Cl_2\rightarrow2HCl\left(1\right)\)
\(HCl+AgNO_3\rightarrow AgCl+HNO_3\left(2\right)\)
Ta có :
\(n_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{Cl2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PTHH: (1)
\(\Rightarrow\frac{0,1}{1}< \frac{0,15}{1}\) nên sau phản ứng H2 hết, Cl2 dư
\(\Rightarrow n_{HCl}=2n_{H2}=0,2\left(mol\right)\)
Theo PTHH (2) \(\Rightarrow n_{AgCl}=0,2\left(mol\right)\)
\(\Rightarrow m_{AgCl}=0,2.143,5=28,7\%\)
\(\Rightarrow H=\frac{11,48}{28,7}.100\%=40\%\)