\(n_{Na_2O}=\dfrac{1.86}{62}=0.03\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(0.03........................0.06\)
\(C_{M_{NaOH}}=\dfrac{0.06}{0.25}=0.24\left(M\right)\)
\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
\(0.06...........0.03\)
\(V_{CO_2}=0.03\cdot22.4=0.672\left(l\right)\)
Sửa $1,68 \to 1,86$
a) $Na_2O + H_2O \to 2NaOH$
b) n Na2O = 1,86/62 = 0,03(mol)
n NaOH = 2n Na2O = 0,06(mol)
=> CM NaOH = 0,06/0,25 = 0,24M
c) $CO_2 + 2NaOH \to Na_2CO_3 + H_2O$
n CO2 = 1/2 n NaOH = 0,03(mol)
=> V CO2 = 0,03.22,4 = 0,672(lít)