\(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\\ n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
0,1----------------->0,2
\(2NaOH+Al_2O_3\rightarrow2NaAlO_2+H_2O\)
bđ 0,2 0,2
pư 0,2 0,1
spư 0 0,1 0,2
Vậy dd sau phản ứng trong dd có các chất tan: \(NaAlO_2\)
\(m_{dd}=500+6,2+0,1.102=516,4\left(g\right)\\ m_{NaAlO_2}=0,2.82=16,4\left(g\right)\\ \rightarrow C\%_{NaAlO_2}=\dfrac{16,4}{516,4}.100\%=3,18\%\)
\(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
0,1 0,2 ( mol )
\(m_{NaOH}=0,2.40=8\left(g\right)\)
\(m_{ddspứ}=6,2+20,4+500=526,6\left(g\right)\)
\(C\%_{NaOH}=\dfrac{8}{526,6}.100=1,51\%\)
\(C\%_{Al_2O_3}=\dfrac{20,4}{526,6}.100=3,87\%\)
