Ta có:
\(n_{Cl2}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
\(NaI+Cl_2\rightarrow2NaCl+I_2\)
\(NaCl+AgNO_3\rightarrow AgCl+NaNO_3\)
\(\Rightarrow n_{NaCl}=2n_{Cl2}=n_{AgNO3}=0,05.2=0,1\left(mol\right)\)
\(\Rightarrow V_{AgNO3}=\frac{0,1}{0,5}=0,2\left(l\right)\)