K/l H2SO4:
......mH2SO4=114.20%=22,8(g)
mBaCl2=400.5,2%=20,8(g)
PTHH: H2SO4+BaCl2--->2HCl+ BaSO4
Có: 22,8/(2+32+16.4)>20,8/(137+35,5.2)=> H2SO4 dư \(13g\)
PTHH: H2SO4+BaCl2--->2HCl+ BaSO4
.................9,8......20,8..........7,3.........23,3(g)
Có: m(dd sau p/ứ)=114+400=514(g)
\(C\%_{H_2SO_4}=\frac{13}{514}.100=\frac{650}{257}\left(\%\right)\)
\(C\%_{HCl}=\frac{7,3}{514}.100=\frac{365}{257}\left(\%\right)\)
\(C\%_{BaSO_4}=\frac{23,3}{514}.100=\frac{1165}{257}\left(\%\right)\)
#Walker
Đúng 0
Bình luận (0)
