+nNa2O = 6,2/62 = 0,1 mol
PT
Na2O + H2O -> 2NaOH
0,1_____________0,2 (mol)
mNaOH = 0,2*40 = 8 g
mdd A = mNa2O + m nước = 6,2 + 73,8 = 80 g
-> C% NaOH (dd A) = 8/80 *100% = 10%
\(n_{Na_2O}=\dfrac{m_{Na_2O}}{M_{Na_2O}}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
\(n_{H_2O}=\dfrac{m_{H_2O}}{M_{H_2O}}=\dfrac{73,8}{18}=4,1\left(mol\right)\)
\(PTHH:Na_2O+H_2O----->2NaOH\)
Theo PTHH: \(\dfrac{n_{Na_2O}}{1}< \dfrac{n_{H_2O}}{1}\) nên H2O dư-->Tính theo Na2O
Theo PTHH:\(n_{NaOH}=2n_{Na_2O}=2.0,1=0,2\left(mol\right)\)
\(m_{NaOH}=n_{NaOH}.M_{NaOH}=0,2.40=8\left(g\right)\)
Ta có \(m_{dd\left(A\right)}=m_{Na_2O}+m_{H_2O}=6,2+73,8=80\left(g\right)\)
\(C\%_{NaOH\left(A\right)}=\dfrac{m_{NaOH}}{m_{dd\left(A\right)}}.100\%=\dfrac{8}{80}.100\%=10\%\)
