Question

Answer 1

Câu 1:

(1) \(4FeS_2+11O_2\underrightarrow{t^o}2Fe_2O_3+8SO_2\)

(2) \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

(3) \(Fe_2\left(SO_4\right)_3+6NaOH\rightarrow2Fe\left(OH\right)_3\downarrow+3Na_2SO_4\)

(4) \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

Câu 2

a) 2K + 2H₂O --> 2KOH + H₂

Ba+ 2H₂O --> Ba(OH)₂ + H₂

b)\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

Gọi số mol K, Ba là a, b

=> 39a + 137b = 7,04

PTHH: 2K + 2H₂O --> 2KOH + H₂

______a---------------------->0,5a

Ba+ 2H₂O --> Ba(OH)₂ + H₂

b--------------------------->b

=> 0,5a + b = 0,06

=> a = 0,04; b = 0,04

=> \(\left\{{}\begin{matrix}\%K=\dfrac{0,04.39}{7,04}.100\%=22,159\%\\\%Ba=\dfrac{0,04.137}{7,04}.100\%=77,841\%\end{matrix}\right.\)