Question

Answer 1

a) \(n_R=\dfrac{5,85}{M_R}\left(mol\right)\)

\(n_{H_2SO_4}=0,15.0,5=0,075\left(mol\right)\)

PTHH: 2R + 2H₂O --> 2ROH + H₂

____\(\dfrac{5,85}{M_R}\)------------>\(\dfrac{5,85}{M_R}\)

2ROH + H₂SO₄ --> R₂SO₄ + H₂O

0,15<-0,075

=> \(\dfrac{5,85}{M_R}=0,15=>M_R=39\left(g/mol\right)\) => R là K(Kali)

b) 

\(n_K=\dfrac{5,85}{39}=0,15\left(mol\right)\)

PTHH: 2K + 2H₂O --> 2KOH + H₂

_____0,15---------------------->0,075

=> V_H2 = 0,075.22,4 = 1,68(l)

c) 

PTHH: 4K + O₂ --t^o--> 2K₂O

_____0,15------------>0,075

=> m_K2O = 0,075.94 = 7,05(g)