Câu 4:
a) \(n_{BaCl_2}=\dfrac{150.31,2\%}{208}=0,225\left(mol\right)\);
\(n_{Na_2CO_3}=\dfrac{150.23,85\%}{106}=0,3375\left(mol\right)\)
PTHH: \(BaCl_2+Na_2CO_3\rightarrow BaCO_3\downarrow+2NaCl\)
Xét tỉ lệ: \(\dfrac{0,225}{1}< \dfrac{0,3375}{1}\) => BaCl2 hết, Na2CO3 dư
PTHH: \(BaCl_2+Na_2CO_3\rightarrow BaCO_3\downarrow+2NaCl\)
0,225----->0,225------>0,225---->0,45
X chứa \(\left\{{}\begin{matrix}NaCl:0,45\left(mol\right)\\Na_2CO_{3\left(dư\right)}:0,3375-0,225=0,1125\left(mol\right)\end{matrix}\right.\)
PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
0,1125-->0,225
=> \(V=\dfrac{0,225}{1}=0,225\left(l\right)=225\left(ml\right)\)
b)
mdd X = 150 + 150 - 0,225.197 = 255,675 (g)
\(\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{0,45.58,5}{255,675}.100\%=10,3\%\\C\%_{Na_2CO_3}=\dfrac{0,1125.106}{255,675}.100\%=4,66\%\end{matrix}\right.\)
