\(AgNO3+NaCl-->NaNO3+AgCl\downarrow\)
\(m_{AgNO3}=\frac{100.17}{100}=17\left(g\right)\)
\(n_{AgNO3}=\frac{17}{170}=0,1\left(mol\right)\)
\(n_{NaCl}=\frac{400.2,94}{100}=11,76\left(g\right)\)
\(n_{NaCl}=\frac{11,76}{58,5}=0,2\left(mol\right)\)
--> NaCl dư
\(n_{AgCl}=n_{AgNO3}=0,1\left(mol\right)\)
\(m_C=m_{AgCl}=0,1.143,5=14,35\left(g\right)\)
dd X gồm NaCl dư và NaNO3
m dd sau pư = mdd AgNO3 + mdd NaCl -m AgCl = 100+400-14,35=485,65(g)
\(n_{NaNO3}=n_{AgNO3}=0,1\left(mol\right)\)
\(m_{NaNO3}=0,1.85=8,5\left(g\right)\)
\(C\%_{NaNO3}=\frac{8,5}{485,65}.100\%=1,75\%\)
\(n_{NaCl}=n_{AgNO3}=0,1\left(Mol\right)\)
\(n_{NaC_{ }l}dư=0,2-0,1=0,1\left(mol\right)\)
\(m_{NaCl}=0,1.58,5=5,85\left(g\right)\)
\(C\%_{NaCl}=\frac{5,85}{485,65}.100\%=1,2\%\)
