NaOH + HCl ➜ NaCl + H2O
\(m_{NaOH}=m_{HCl}=10\times20\%=2\left(g\right)\)
\(\Rightarrow n_{NaOH}=\dfrac{2}{40}=0,05\left(mol\right)\)
\(n_{HCl}=\dfrac{2}{36,5}=\dfrac{4}{73}\left(mol\right)\)
Theo PT: \(n_{NaOH}=n_{HCl}\)
Theo bài: \(n_{NaOH}=\dfrac{73}{80}n_{HCl}\)
Vì \(\dfrac{73}{80}< 1\Rightarrow\) NaOH hết, HCl dư
Dung dịch sau phản ứng gồm: HCl dư và NaCl
Do trong dung dịch có axit nên làm màu quỳ tìm chuyển đỏ.
b) Theo PT: nNaCl=nNaOH=0,05(mol)
\(\Rightarrow m_{NaCl}=0,05\times58,5=2,925\left(g\right)\)
\(m_{dd}=10+10=20\left(g\right)\)
Theo PT: \(n_{HCl}pư=n_{NaOH}=0,05\left(mol\right)\)
\(\Rightarrow n_{HCl}dư=\dfrac{4}{73}-0,05=\dfrac{7}{1460}\left(mol\right)\)
\(\Rightarrow m_{HCl}dư=\dfrac{7}{1460}\times36,5=0,175\left(g\right)\)
\(\Rightarrow C\%_{ddNaCl}=\dfrac{2,925}{20}\times100\%=14,625\%\)
\(C\%_{ddHCl_{dư}}=\dfrac{0,175}{20}\times100\%=0,875\%\)