\(\text{a. PTHH: BaCl2 + H2SO4 ->BaSO4 + 2HCl}\)
\(\text{b.n BaCl2= mBaCl2 / MBaCl2 = 20,8/208=0,1(mol)}\)
\(\text{PT: BaCl2 + H2SO4 -> BaSO4 + 2HCl}\)
Tỉ lệ: 1....................1................1...............2...........(mol)
P/ư: 0,1.................0,1...............0,1..................0,2 ...........(mol)
Từ pt=> n BaSO4 = 0,1 mol
\(\text{mBaSO4= nBaSo4 . M BaSO4 = 0,1 . 233 =23,4 (g)}\)
c. Từ pt => n H2SO4 =0,1 mol
mH2S04 = nH2SO4 . M H2S04 = 0,1.98 =9,8(g)
\(\text{m ddH2SO4 = (m H2SO4 . 100)/20= 49g}\)
d.
V ddH2SO4= mddH2SO4 / D H2SO4 = 49 / 1,14 =42,98(ml)=0,04298 l
\(\Rightarrow\text{CM H2S04 = m H2S04 / VddH2SO4 = 49/0,04298=1140 M}\)