\(n_{BaCl_2}=\dfrac{8,32}{208}=0,04\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{200.20\%}{98}=0,41\left(mol\right)\)
Pt: \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
0,04 0,41 -------> 0,04mol--> 0,08mol
Lập tỉ số: \(n_{BaCl_2}:n_{H_2SO_4}=0,04< 0,41\)
BaCl2 hết, H2SO4 dư
\(n_{H_2SO_4\left(dư\right)}=0,41-0,04=0,37\left(mol\right)\)
\(\Sigma_{m_{dd\left(spu\right)}}=8,32+200-0,04.233=199\left(g\right)\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,37.98.100}{199}=18,22\%\)
\(C\%_{HCl}=\dfrac{0,08.36,5.100}{199}=1,47\%\)
Theo đề bài ta có : \(\left\{{}\begin{matrix}nBaCl2=\dfrac{8,32}{208}=0,04\left(mol\right)\\nH2SO4=\dfrac{200.20}{100.98}\approx0,4\left(mol\right)\end{matrix}\right.\)
Ta có pTHH :
\(BaCl2+H2So4\rightarrow B\text{aS}O4\downarrow+2HCl\)
0,04mol........0,04mol........................0,08mol
Theo PTHH ta có : \(nBaCl2=\dfrac{0,04}{1}mol< nH2SO4=\dfrac{0,4}{1}mol\) => nH2SO4 dư ( tính theo nBaCl2)
Ta có : \(\left\{{}\begin{matrix}C\%H2SO4\left(d\text{ư}\right)=\dfrac{\left(0,4-0,04\right).98}{8,32+200}.100\%\approx16,94\%\\C\%_{HCl}=\dfrac{0,08.36,5}{8,32+200}.100\%\approx1,40\%\end{matrix}\right.\)
