Gọi số mol Na2O, Al2O3(bđ) là a, b (mol)
\(n_{Al_2O_3\left(pư\right)}=b-\dfrac{10,2}{102}=b-0,1\left(mol\right)\)
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
a---------------->2a
\(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)
a<--------2a---------->2a
\(\left\{{}\begin{matrix}n_{Al_2O_3\left(pư\right)}=a=b-0,1\\m_{NaAlO_2}=164a\left(g\right)\end{matrix}\right.\)
mdd sau pư = 62a + 102a + 200
= 164a + 200
Ta có: \(C\%_{NaAlO_2}=\dfrac{164a}{164a+200}.100\%=10\%\)
=> a = \(\dfrac{50}{369}\left(mol\right)\Rightarrow b=\dfrac{869}{3690}\left(mol\right)\)
\(\left\{{}\begin{matrix}\%m_{Na_2O}=\dfrac{\dfrac{50}{369}.62}{\dfrac{50}{369}.62+\dfrac{869}{3690}.102}.100\%=25,91\%\\\%m_{Al_2O_3}=\dfrac{\dfrac{869}{3690}.102}{\dfrac{869}{3690}.102+\dfrac{50}{369}.62}.100\%=74,09\%\end{matrix}\right.\)
