a) \(n_{AgNO_3}=\dfrac{200.34\%}{170}=0,4\left(mol\right)\)
PTHH: \(AgNO_3+NaCl\rightarrow AgCl\downarrow+NaNO_3\)
0,4-------->0,4----->0,4------->0,4
b) \(m_{kt}=0,4.143,5=57,4\left(g\right)\)
c) \(m_{ddNaCl}=\dfrac{0,4.58,5}{25\%}=93,6\left(g\right)\)
d) \(m_{dd.sau.pư}=200+93,6-236,2\left(g\right)\)
=> \(C\%_{NaNO_3}=\dfrac{0,4.85}{236,2}.100\%=14,4\%\)