mBaCl2=250.10%=25(g) => nBaCl2= 25/208 (mol)
mH2SO4=40.15%= 6(g) => nH2SO4= 6/98=3/49(mol)
PTHH: BaCl2 + H2SO4 -> BaSO4 + 2 HCl
Ta có: 25/208 > 3/49
-> H2SO4 hết, BaCl2 dư => Tính theo nH2SO4
a) nBaSO4=nBaCl2(p.ứ)=nH2SO4=3/49(mol)
=>m(kết tủa)= mBaSO4=233. 3/49= 14,265(g)
b) nBaCl2(dư)= 25/208 - 3/49= 601/10192
=> mBaCl2(dư)= 601/10192 . 208= 12,265(g)
nHCl= 2. 3/49 = 6/49 (mol) => mHCl=6/49. 36,5= 4,469(g)
mddsau= 250+ 40 - 14,265= 275,735(g)
=>C%ddBaCl2(dư)= (12,265/275,735).100=4,448%
C%ddHCl= (4,469/275,735).100=1,621%
Đúng 1
Bình luận (0)
