\(m_{NaCl}=\dfrac{5,85.100}{100}=5,85g\)
\(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\)
\(m_{AgNO_3}=\dfrac{12.100}{100}=12g\)
\(n_{AgNO_3}=\dfrac{12}{170}\approx0,07\left(mol\right)\)
Ta co: \(\dfrac{0,1}{1}>\dfrac{0,07}{1}\Rightarrow\) NaCl dư
NaCl + AgNO3 \(\rightarrow\) NaNO3 + AgCl\(\left(\downarrow\right)\)
de: 0,1 0,07
pu: 0,07 0,07 0,07 0,07
spu: 0,03 0 0,07 0,07
\(m_{NaCl\left(dư\right)}=0,03.58,5=1,755g\)
\(m_{NaNO_3}=0,07.85=5,95g\)
\(m_{ddspu}=m_{ddNaCl}+m_{ddAgNO_3}-m_{AgCl}=189,955g\)
\(C\%_{NaCl\left(dư\right)}=\dfrac{1,755}{189,955}.100\%\approx0,92\%\)
\(C\%_{NaNO_3}=\dfrac{5,95}{189,955}.100\%\approx3,13\%\)
