Câu 2:
Gọi \(\left\{{}\begin{matrix}n_{Na_2CO_3}=x\left(mol\right)\\n_{Na_2SO_4}=y\left(mol\right)\end{matrix}\right.\)
PTHH:
\(Na_2CO_3+BaCl_2\rightarrow BaCO_3\downarrow+2NaCl\)
\(Na_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2NaCl\)
\(\xrightarrow[]{BT\text{ }SO_4^{2-},CO_3^{2-}}\left\{{}\begin{matrix}n_{BaCO_3}=x\left(mol\right)\\n_{BaSO_4}=y\left(mol\right)\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}106x+142y=\dfrac{6,63}{1,7}=3,9\\197x+233y=6,63\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,01\\y=0,02\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}C_{M\left(Na_2CO_3\right)}=\dfrac{0,01}{0,25}=0,04M\\C_{M\left(Na_2SO_4\right)}=\dfrac{0,02}{0,25}=0,08M\end{matrix}\right.\)
Câu 3:
Gọi \(\left\{{}\begin{matrix}n_{NaOH}=x\left(mol\right)\\n_{Na_2CO_3}=y\left(mol\right)\\n_{Na_2SO_4}=z\left(mol\right)\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}n_{HCl}=0,2.0,08=0,016\left(mol\right)\\n_{BaCl_2}=\dfrac{4,16}{208}=0,02\left(mol\right)\end{matrix}\right.\)
PTHH:
\(NaOH+HCl\rightarrow NaCl+H_2O\\ Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\\ Na_2CO_3+BaCl_2\rightarrow BaCO_3\downarrow+2NaCl\\ Na_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2NaCl\)
\(\xrightarrow[]{BT\text{ }Na^+,Cl^-,CO_3^{2-},SO_4^{2-}}\left\{{}\begin{matrix}x+2y=0,016\\y+z=0,02\\208y+233z=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,0016\\y=0,0072\\z=0,0128\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}C\%_{NaOH}=\dfrac{0,0016.40}{80}.100\%=0,08\%\\C\%_{Na_2CO_3}=\dfrac{0,0072.106}{80}.100\%=0,954\%\\C\%_{Na_2SO_4}=\dfrac{0,0128.142}{80}.100\%=2,272\%\end{matrix}\right.\)
