nBa(OH)2= 0.15*1 = 0.15 nNa2CO3 dư = 0.2 - 0.15 = 0.05 mNaOH = 0.3*40 = 12 mNa2CO3 dư = 0.05*106 = 5.3 = 110 + 168 - 29.55 = 248.45 a. C%NaOH = 12/248.45*100 = 4.8% C%Na2CO3 dư = 5.3/248.45*100 = 2.13% Vd d = VNa2CO3 + VBa(OH)2 = 0.1 + 0.15 = 0.25 CM[NaOH] = 0.3/0.25 = 1.2M CM[Na2CO3] = 0.05/0.25 = 0.2M b. nHCl = 0.3 mHCl = 0.3*36.5 = 10.95 mddHCl = 10.95*100/7.3 = 150 VHCl = 150/1.08 = 138ml c. nCO2 = 0.15 VCO2 = 0.15*22.4 = 3.36L
Na2CO3 + Ba(OH)2 ------> BaCO3 + 2NaOH 0.15------------ 0.15------------- 0.15-------- 0.3 Dung dịch A gồm NaOH, Na2CO3 Kết tủa C là BaCO3 cho tác dụng vs HCl BaCO3 + 2HCl -------> BaCl2 + CO2 + H2O 0.15--------- 0.3------------------------ 0.15 nNa2CO3 bđ = 0.1*2 = 0.2 nBa(OH)2= 0.15*1 = 0.15 nNa2CO3 dư = 0.2 - 0.15 = 0.05 mNaOH = 0.3*40 = 12 mNa2CO3 dư = 0.05*106 = 5.3
mdd sau pư = mddNa2CO3 + mddBa(OH)2 - mBaCO3 = (100*1.1) + (150*1.12) - (197*0.15) = 110 + 168 - 29.55 = 248.45 a. C%NaOH = 12/248.45*100 = 4.8% C%Na2CO3 dư = 5.3/248.45*100 = 2.13% Vd d = VNa2CO3 + VBa(OH)2 = 0.1 + 0.15 = 0.25 CM[NaOH] = 0.3/0.25 = 1.2M CM[Na2CO3] = 0.05/0.25 = 0.2M b. nHCl = 0.3 mHCl = 0.3*36.5 = 10.95 mddHCl = 10.95*100/7.3 = 150 VHCl = 150/1.08 = 138ml c. nCO2 = 0.15 VCO2 = 0.15*22.4 = 3.36L