mNa2CO3 = 10.6 g
nNa2CO3 = 0.1 mol
mHCl = 100*7.3/100= 73g
nHCl = 2 mol
Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O
Bđ: 0.1________2
Pư : 0.1________0.2_____0.2____0.1
Kt : 0_________1.8______0.2____0.1
VCO2 = 2.24 l
mCO2 = 4.4g
mHCl dư = 65.7 g
mNaCl = 11.7 g
mdd sau phản ứng = 200 + 100 - 4.4 = 295.6 g
C%HCl dư = 22.22%
C%NaCl = 3.95%
OH- + H+ --> H2O
1.8___1.8
CM NaOH = 1.8/0.1 = 18M