Ta có: \(n_{KHCO_3}=\dfrac{20.10\%}{100}=0,02\left(mol\right)\)
\(n_{HCl}=\dfrac{50.14,6\%}{36,5}=0,2\left(mol\right)\)
PT: \(KHCO_3+HCl\rightarrow KCl+CO_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,02}{1}< \dfrac{0,2}{1}\), ta được HCl dư.
Theo PT: \(n_{HCl\left(pư\right)}=n_{KCl}=n_{CO_2}=n_{KHCO_3}=0,02\left(mol\right)\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,2-0,02=0,18\left(mol\right)\)
Ta có: m dd sau pư = 20 + 50 - 0,02.44 = 69,12 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{KCl}=\dfrac{0,02.74,5}{69,12}.100\%\approx2,16\%\\C\%_{HCl}=\dfrac{0,18.36,5}{69,12}.100\%\approx9,51\%\end{matrix}\right.\)
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