\(n_{CaCl_2}=\dfrac{22.2}{111}=0.2\left(mol\right)\)
\(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\)
\(0.2....................................................0.4\)
\(m_{AgCl}=0.4\cdot143.5=57.4\left(g\right)\)
nCaCl2=22,2/111=0,2(mol)
CaCl2 + 2AgNO3 -----> 2AgCl + Ca(NO3)2
TPT:nAgCl=2.nCaCl2=2.0,2=0,4(mol)
mAgCl=0,4.143,5=57,4(g)