Question

Answer 1

Fe2O3 + 6HCl -> 2FeCl3 + 3H2O

nFe2O3 = 0,25 

nHCl = 0,6 

\(\dfrac{0,25}{1}>\dfrac{0,6}{6}\)=> Fe2O3 dư 0,15 mol => m Fe dư = 24g

theo pư => n FeCl3 = 0,2 mol => m FeCl3 = 32,5 g

 

 

Answer 2

Câu 5 : 

\(n_{Al_2O_3}=\dfrac{78}{102}=\dfrac{13}{17}\left(mol\right)\)

\(Al_2O_3+6HNO_3\rightarrow2Al\left(NO_3\right)_3+3H_2O\)

\(\dfrac{13}{17}.......\dfrac{78}{17}........\dfrac{26}{17}\)

\(V_{dd_{HNO_3}}=\dfrac{\dfrac{78}{17}}{2}=2.3\left(l\right)\)

\(C_{M_{Al\left(NO_3\right)_3}}=\dfrac{\dfrac{26}{17}}{2.3}=0.6\left(M\right)\)