PTHH: Na2CO3 + CaCl2 -> 2NaCl + CaCO3
mNa2CO3 = 265.10% = 26,5 (g)
=> n Na2CO3 = 0,25 mol
mCaCl2 = 500.6,6% = 33 (g)
=> nCaCl2 = 11/37 mol
Vì 0,25/1< (11/37)/1 => sau pứ CaCl2 dư
=> nCaCO3=nNa2CO3=0,25 mol
=>mdd sau pứ=265 + 500 - 0,25.100 =740g
nCaCl2 pứ=nNa2CO3=0,25 mol
=> nCaCl2 dư=11/37-0,25= 7/148 mol
C% dd CaCl2 = (7/148).111/740 .100% = 0,71%
C% dd NaCl = 0,5.58,5/740 .100% = 3,95%