\(n_{CO_2}=\frac{2,016}{22,4}=0,09\left(mol\right)\)
Gọi a,b lần lượt là số mol của Na2CO3 và K2CO3
Na2CO3 + 2HCl -> 2NaCl + CO2↑ + H2O
a...............................2a..............a
K2CO3 + 2HCl -> 2KCl + CO2↑ + H2O
b...............................2b................b
Ta có hệ pt: \(\left\{{}\begin{matrix}106a+138b=10,5\\a+b=0,09\end{matrix}\right.\)
Giải hệ ta được: a = 0,06 ; b = 0,03
=> \(m_{Na_2CO_3}=106.0,06=6,36\left(g\right)\)
\(m_{K_2CO_3}=138.0,03=4,14\left(g\right)\)
=>\(\%m_{Na_2CO_3}=\frac{6,36}{10,5}\cdot100\%=60,57\%\)
\(\%m_{K_2CO_3}=100\%-60,57\%=39,43\%\)
Theo pthh ta có: \(n_{NaCl}=2n_{Na_2CO_3}=2.0,06=0,12mol;n_{KCl}=2n_{K_2CO_3}=2.0,03=0,06mol\)
=>\(m_{NaCl}=0,12.58,5=7,02\left(g\right)\)
\(m_{KCl}=0,06.74,5=4,47\left(g\right)\)
a) Na2CO3+2HCl---->2NaCl+H2O+CO2
x-----------------------------------------------x
K2CO3+2HCl---->2KCl+H2O+CO2
y--------------------------------------y
b) Gọi n\(_{Na2CO3}=x\Rightarrow m_{Na2CO3}=106x\)
n\(_{K2CO3}=y\Rightarrow m_{K2CO3}=138y\)
n\(_{CO2}=\frac{2,016}{22,4}=0,09\left(mol\right)\)
Theo bài ra ta có
\(\left\{{}\begin{matrix}106x+138y=10,5\\x+y=0,09\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,06\\y=0,03\end{matrix}\right.\)
%m\(_{Na2CO3}=\frac{0,06.106}{10,5}.100\%=60,57\%\)
%m\(_{K2CO3}=100-60,57=39,43\%\)
c) Theo pthh
n\(_{KCl}=2n_{K2CO3}=0,06\left(mol\right)\)
m\(_{KCl}=0,06.74,5=4,47\left(g\right)\)
Theo pthh
n\(_{ }\)\(_{NaCl}=2n_{Na2CO3}=1,2\left(mol\right)\)
m\(_{NaCl}=1,2.58,5=70,2\left(g\right)\)
\(\)
