Дано
m(H2SO4)=20 g
m(ppa BaCL2)=150 g
W=10%
-----------------------------
m(BaSO4)-?
m(в-ва BaCL2)=150*10%/100%=15 g
20 15 X
H2SO4+BaCL2-->2HCL+BaSO4
98 208 233
M(H2SO4)=98 g/mol M(BaCL2)=208 g/mol M(BaSO4)=233 g/mol
n(H2SO4)=m/M=20/98=0.2 mol
n(BaCL2)=15/208=0.07 mol
n(H2SO4)>n(BaCL2)
15/208 = X/233
X=16.8 g
ответ 16.8 г
Mg2+,Kk+, H+, Al3+ - Катионы
Анионы - O2-, S2-, F-, N3-
Zn + 2HCl = ZnCl2 + H2
m(HCl) = 80*20 /100 = 16 г
M(HCl) = 36.5 г/моль
n(HCl) = 16/36.5 = 0.44 моль
n(H2) = 0.44 :2 = 0.22 моль
V(H2) = 22.4 * 0.22 = 4.93 л
1)AlCl3+3NaOH=Al(OH)3+3NaCl
+3 -1 +1 -1 +1 -1
Al+3Cl+3Na+3OH=Al(OH)3+3Na+3CL
+3 -1
Al+3OH=Al(OH)3
0 +1-1 +2-1 0
2) Mg+2HCl=MgCl2+H2
0 +1 +2 0
Mg+2H=Mg+H2