a, Ta có: 100nKHCO3 + 126nNa2SO3 = 16,3 (1)
PT: \(HCl+KHCO_3\rightarrow KCl+CO_2+H_2O\)
\(2HCl+Na_2SO_3\rightarrow2NaCl+SO_2+H_2O\)
Theo PT: \(n_{CO_2}+n_{SO_2}=n_{KHCO_3}+n_{Na_2SO_3}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{KHCO_3}=0,1\left(mol\right)\\n_{Na_2SO_3}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{KHCO_3}=\dfrac{0,1.100}{16,3}.100\%\approx61,3\%\\\%m_{Na_2SO_3}\approx38,7\%\end{matrix}\right.\)
b, Theo PT: \(n_{HCl\left(pư\right)}=n_{KHCO_3}+2n_{Na_2SO_3}=0,2\left(mol\right)\)
Mà HCl dùng dư 10%
⇒ nHCl = 0,2 + 0,2.10% = 0,22 (mol)
\(\Rightarrow V_{ddHCl}=\dfrac{0,22}{1}=0,22\left(l\right)\)
B gồm: KCl, NaCl và HCl dư.
Có: nHCl (dư) = 0,2.10% = 0,02 (mol)
Theo PT: \(\left\{{}\begin{matrix}n_{KCl}=n_{KHCO_3}=0,1\left(mol\right)\\n_{NaCl}=2n_{Na_2SO_3}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{KCl}}=C_{M_{NaCl}}=\dfrac{0,1}{0,22}=0,45\left(M\right)\\C_{M_{HCl}}=\dfrac{0,02}{0,22}=0,09\left(M\right)\end{matrix}\right.\)
c,\(n_{SO_2}=n_{Na_2SO_3}=0,05\left(mol\right)\)
PT: \(SO_2+Br_2+2H_2O\rightarrow H_2SO_4+2HBr\)
_____0,05____0,05 (mol)
\(\Rightarrow m_{ddBr_2}=\dfrac{0,05.160}{3,2\%}=250\left(g\right)\)