Na2CO3 + 2HCl → 2NaCl + CO2 + H2O (1)
NaHCO3 + HCl → NaCl + CO2 + H2O (2)
\(n_{CO_2}=\frac{0,896}{22,4}=0,04\left(mol\right)\)
a) Gọi x,y lần lượt là số mol của Na2CO3 và NaHCO3
Theo PT1,2 ta có: \(\left\{{}\begin{matrix}106x+84y=3,8\\x+y=0,04\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,02\end{matrix}\right.\)
\(n_{Na_2CO_3}=0,02\left(mol\right)\Rightarrow m_{Na_2CO_3}=0,02\times106=2,12\left(g\right)\)
\(n_{NaHCO_3}=0,02\left(mol\right)\Rightarrow m_{NaHCO_3}=0,02\times84=1,68\left(g\right)\)
\(\%m_{Na_2CO_3}=\frac{2,12}{3,8}\times100\%=55,79\%\)
\(\%m_{NaHCO_3}=100\%-55,79\%=44,21\%\)
b) Theo PT1: \(n_{HCl}=2n_{Na_2CO_3}=2\times0,02=0,04\left(mol\right)\)
Theo PT2: \(n_{HCl}=n_{NaHCO_3}=0,02\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,04+0,02=0,06\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,06\times36,5=2,19\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{2,19}{20\%}=10,95\left(g\right)\)
\(\Rightarrow V_{ddHCl}=\frac{10,95}{1,1}=9,95\left(ml\right)\)
c) Theo PT1: \(n_{NaCl}=n_{HCl}=0,04\left(mol\right)\)
Theo pT2: \(n_{NaCl}=n_{HCl}=0,02\left(mol\right)\)
\(\Rightarrow\Sigma n_{NaCl}=0,02+0,04=0,06\left(mol\right)\)
\(m_{NaCl}=0,06\times58,5=3,51\left(g\right)\)
\(C_{M_{NaCl}}=\frac{0,06}{0,0095}=6,32\left(M\right)\)
\(m_{CO_2}=0,04\times44=1,76\left(g\right)\)
Ta có: \(m_{dd}saupư=3,8+10,95-1,76=12,99\left(g\right)\)
\(C\%_{NaCl}=\frac{3,51}{12,99}\times100\%=27,02\%\)
