a,
V dd HCl= 220/1,1= 200 cm3= 0,2l
=> nHCl= 0,2.0,9= 0,18 mol
Na2CO3+ 2HCl -> 2NaCl+ CO2+ H2O
K2CO3+ 2HCl -> 2KCl+ CO2+ H2O
Gọi x là mol Na2CO3, y là mol K2CO3
Ta có hệ:
106x+ 138y= 10,82
2x+2y= 0,18
<=> x= 0,05; y= 0,04
%Na2CO3= $\frac{0,05.106.100}{10,82}$= 49%
%K2CO3= 51%
b,
nCO2= x+y= 0,09 mol
Ta có $n= \frac{PV}{RT}$
<=> $0,09= \frac{2V}{0,082.(27+273)}$
<=> $V= 1,107l$
c,
nNaCl= 2x= 0,1 mol => mNaCl= 5,85g
nK= 2y= 0,08 mol => mKCl= 5,96g
mCO2= 0,09.44= 3,96g
m dd spu= 10,82+220-3,96= 226,86g
C%NaCl= $\frac{5,85.100}{226,86}$= 2,58%
C%KCl= $\frac{5,96.100}{226,86}$= 2,63%