\(n_{H_2SO_4}=\dfrac{500.4,9\%}{98}=0,25\left(mol\right)\)
\(n_{Fe_2\left(SO_4\right)_3}=\dfrac{500.20\%}{400}=0,25\left(mol\right)\)
Gọi số mol Na là k (mol)
\(n_{Fe\left(OH\right)_3}=\dfrac{32,1}{107}=0,3\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
k---------------->k------>0,5k
2NaOH + H2SO4 --> Na2SO4 + 2H2O
0,5<-----0,25------->0,25
6NaOH + Fe2(SO4)3 --> 3Na2SO4 + 2Fe(OH)3
0,9<------0,15<---------0,45<--------0,3
=> k = 0,5 + 0,9 = 1,4 (mol)
=> m = 1,4.23 = 32,2 (g)
mdd sau pư = 32,2 + 500 - 0,7.2 - 32,1 = 498,7 (g)
\(\left\{{}\begin{matrix}n_{Na_2SO_4}=0,7\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,1\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C\%_{Na_2SO_4}=\dfrac{0,7.142}{498,7}.100\%=19,93\%\\C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{498,7}.100\%=8,02\%\end{matrix}\right.\)
