Đổi 600ml = 0,6 lít
Ta có: \(n_{AgNO_3}=2.0,6=1,2\left(mol\right)\)
PTHH: AgNO3 + NaCl ---> AgCl + NaNO3
Theo PT: \(n_{NaCl}=n_{AgCl}=1,2\left(mol\right)\)
=> \(V_{dd_{NaCl}}=\dfrac{1,2}{1,5}=0,8\left(lít\right)=800ml\)
=> \(m_{AgCl}=1,2.143,5=172,2\left(g\right)\)