a, PT: \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_{4\downarrow}\)
b, Ta có: \(n_{BaCl_2}=\dfrac{312.20\%}{208}=0,3\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{196.10\%}{98}=0,2\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\), ta được BaCl2 dư.
Theo PT: \(n_{BaSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=0,2.233=46,6\left(g\right)\)
c, Theo PT: \(\left\{{}\begin{matrix}n_{BaCl_2\left(pư\right)}=n_{H_2SO_4}=0,2\left(mol\right)\\n_{HCl}=2n_{H_2SO_4}=0,4\left(mol\right)\end{matrix}\right.\)
⇒ nBaCl2( dư) = 0,1 (mol)
Ta có: m dd sau pư = 312 + 196 - 46,6 = 461,4 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{BaCl_2\left(dư\right)}=\dfrac{0,1.208}{461,4}.100\%\approx4,51\%\\C\%_{HCl}=\dfrac{0,4.36,5}{461,4}.100\%\approx3,16\%\end{matrix}\right.\)
