PTHH:2NaOH+CO2-----Na2CO3+H2O
\(V_{CO_2}=448ml=0,448\left(l\right)\)
\(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
Theo PTHH:\(n_{NaOH}=2n_{CO_2}=2.0,02=0,04\left(mol\right)\)
\(m_{NaOH}=n_{NaOH}.M_{NaOH}=0,04.40=1,6\left(g\right)\)
PTHH :
2NaOH + CO2 \(\rightarrow\) Na2CO3 + H2O
ta có : VCO2 =448ml = 0,448 lít
=> nCO2 = \(\dfrac{V_{CO2}}{22,4}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
từ PTHH ta có :
nNaOH = 0,04(mol)
=> mNaOH = nNaOH . MNaOH = 0,04. 20 = 1,6 (g)