PTHH: \(Na_2SO_4+BaCl_2\rightarrow2NaCl+BaSO_4\downarrow\)
a) Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_4}=\frac{200\cdot10\%}{142}=\frac{10}{71}\left(mol\right)\\n_{BaCl_2}=\frac{200\cdot5\%}{208}=\frac{5}{104}\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\) BaCl2 phản ứng hết, Na2SO4 còn dư
\(\Rightarrow n_{BaSO_4}=\frac{5}{104}\left(mol\right)\) \(\Rightarrow m_{BaSO_4}=\frac{5}{104}\cdot233\approx11,2\left(g\right)\)
b)Theo PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=2n_{BaCl_2}=\frac{5}{52}\left(mol\right)\\n_{Na_2SO_4\left(dư\right)}=\frac{685}{7384}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=\frac{5}{52}\cdot58,5=5,625\left(g\right)\\m_{Na_2SO_4\left(dư\right)}=\frac{685}{7384}\cdot142\approx13,17\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddNa_2SO_4}+m_{ddBaCl_2}-m_{BaSO_4}=388,8\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\frac{5,625}{388,8}\cdot100\approx1,45\%\\C\%_{Na_2SO_4}=\frac{13,17}{388,8}\cdot100\approx3,39\%\end{matrix}\right.\)
