\(n_{NaOH}=\dfrac{400.30\%}{40}=3\left(mol\right)\)
\(n_{HCl}=\dfrac{1,14.200.20\%}{36,5}=1,25\left(mol\right)\)
PTHH: NaOH + HCl ----------> NaCl +H2O
Theo đề : 3.........1,25
Lập tỉ lệ :\(\dfrac{3}{1}>\dfrac{1,25}{1}\)=> Sau phản ứng NaOH dư, HCl phản ứng hết
Vậy các dung dịch sau phản ứng là NaOH dư và NaCl
Ta có : \(n_{NaCl}=n_{HCl}=1,25\left(mol\right)\)
\(n_{NaOHdư}=3-1,25=1,75\left(mol\right)\)
\(m_{ddsaupu}=400+1,14.200=628\left(g\right)\)
\(C\%_{NaOHdư}=\dfrac{1,75.40}{628}.100=11,15\%\)
\(C\%_{NaCl}=\dfrac{1,25.58,5}{628}.100=11,64\%\)
Ta có: \(m_{NaOH}=400.30\%=120\left(g\right)\Rightarrow n_{NaOH}=\dfrac{120}{40}=3\left(mol\right)\)
m dd HCl = 200.1,14 = 228 (g)
\(\Rightarrow m_{HCl}=228.20\%=45,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{45,6}{36,5}=\dfrac{456}{365}\left(mol\right)\)
PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Xét tỉ lệ: \(\dfrac{3}{1}>\dfrac{\dfrac{456}{365}}{1}\), ta được NaOH dư.
Theo PT: \(n_{NaOH\left(pư\right)}=n_{NaCl}=n_{HCl}=\dfrac{456}{365}\left(mol\right)\)
\(\Rightarrow n_{NaOH\left(dư\right)}=\dfrac{639}{365}\left(mol\right)\)
Ta có: m dd sau pư = 400 + 228 = 628 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{\dfrac{456}{365}.58,5}{628}.100\%\approx11,64\%\\C\%_{NaOH\left(dư\right)}=\dfrac{\dfrac{639}{365}.40}{628}.100\%\approx11,15\%\end{matrix}\right.\)
Bạn tham khảo nhé!
