\(n_{BaCl_2}=\dfrac{150.31,2}{100}:208=0,225\left(mol\right)\)
\(n_{Na_2CO_3}=\dfrac{150.23,85}{100}:106=0,3375\left(mol\right)\)
\(Na_2CO_3+BaCl_2\rightarrow BaCO_3+2NaCl\)
0,225 0,225 0,225 0,45
Lập tỉ lệ: \(\dfrac{0,225}{1}< \dfrac{0,3375}{1}\)
=> \(Na_2CO_3\) dư sau pứ
dd X: \(\left\{{}\begin{matrix}NaCl\\Na_2CO_3dư\end{matrix}\right.\)
\(kt.Y:BaCO_3\)
\(NaCl.khong.td.với.HCl\)
\(n_{Na_2CO_{3\left(trong.X\right)}}=0,3375-0,225=0,1125\left(mol\right)\)
\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
0,1125 0,225
\(V_{HCl}=\dfrac{0,225}{1}=0,225\left(l\right)\)
\(m_{dd.X}=150+150+0,225.197=344,325\left(g\right)\)
\(C\%_{NaCl}=\dfrac{0,45.58,5.100}{344,325}=7,65\%\)
\(C\%_{Na_2CO_3}=\dfrac{0,1125.106.100}{344,325}=3,46\%\)