PTHH: CO2 + 2NaOH \(\rightarrow\)Na2CO3 + H2O
Theo pt: .. 1 ....... 2 ................ 1 ......... 1 ..... (mol)
Theo đề: . 0,8 .... 1,6 ............ 0,8 ...... 0,8 ... (mol)
\(n_{CO_2}=\dfrac{V_{đktc}}{22,4}=\dfrac{26,88}{22,4}=1,2\left(mol\right)\)
Đổi: 1l = 1000ml
\(m_{ddNaOH}=V_{dd}.D=1000.1,28=1280\left(g\right)\)
\(n_{NaOH}=\dfrac{m_{ddNaOH}.C\%}{100\%.M}=\dfrac{1280.5\%}{100\%.40}=1,6\left(mol\right)\)
So sánh \(1,2>\dfrac{1,6}{2}\left(1,2>0,8\right)\)
=> CO2 dư, tính theo nNaOH
mNa2CO3 = n.M = 0,8.106 = 84,8(g)
mCO2 = n.M = 1,2.44 = 52,8 (g)
mddsau pứ = mCO2 + mdd NaOH = 52,8 + 1280 = 1332,8 (g)
C%Na2CO3 = \(\dfrac{m_{Na_2CO_3}}{m_{ddspứ}}.100\%=\dfrac{84,8}{1332,8}.100\%\approx6,36\%\)
