a) \(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\)
b) \(n_{CaCl_2}=0,2.1=0,2\left(mol\right)\)
=> \(n_{AgCl}=2n_{CaCl2}=0,4\left(mol\right)\)
=> \(m_{AgCl}=0,4.143,5=57,4\left(g\right)\)
c) \(n_{AgNO_3}=2n_{CaCl2}=0,4\left(mol\right)\)
=> \(CM_{AgNO_3}=\dfrac{0,4}{0,4}=1M\)