BaCl2+H2SO4---->BaSO4+2HCl
a) n\(_{BaCl2}=\frac{200.10}{100}=20\left(g\right)\)
n\(_{BaCl2}=\frac{20}{208}=0,096\left(mol\right)\)
Theo pthh
n\(_{HCl}=2n_{BaC_{ }l2}=0,192\left(mol\right)\)
C%=\(\frac{0,192.36,5}{200}.100\%=3,05\%\)
b) n\(_{H2SO4}=n_{BaCl2}=0,096\left(mol\right)\)
m\(_{ddH2SO4}=\frac{0,096.98.100}{15}=62,72\left(g\right)\)
\(\text{mBaCl2=200.10/100=20g}\)
\(\text{⇒nBaCl2=20/208 ≈ 0,09 mol}\)
\(\text{BaCl2 + H2SO4 → BaSO4 + 2HCl}\)
\(\text{0,09 → 0,09 → 0,09 → 0,18}\)
a) mH2SO4=n.M=0,09.98=8,82g
\(\Rightarrow\text{mdd H2SO4=8,82.100/15=58,8g}\)
b) mHCl=n.M=0,18.36,5=6,57g
\(\text{mdd sau pư= mBaCl2+ mH2SO4-mBaSO4}\)
\(\text{= 200+ 58,8-0,09.233}\)
\(\text{= 237,83g}\)
\(\text{⇒ C% HCl=(6,57/237,83).100=2,76%}\)
\(\text{Đổi vị trí hai câu lại mới tính được nhé}\)
