Question

Answer 1

Câu 8: B

M_A2O3 = 2.M_A + 3.16 = 152 (g/mol)

=> M_A = 52 (g/mol)

=> A là Cr

Câu 9: C

 \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PTHH: 2Mg + O₂ --t^o--> 2MgO

            0,2-->0,1

=> V_O2 = 0,1.22,4 = 2,24 (l)

Câu 10: D

Câu 11: B

\(n_R=\dfrac{12,8}{M_R}\left(mol\right)\)

PTHH: 2R + O₂ --t^o--> 2RO

         \(\dfrac{12,8}{M_R}\)----------->\(\dfrac{12,8}{M_R}\)

=> \(\dfrac{12,8}{M_R}\left(M_R+16\right)=16\)

=> M_R = 64 (g/mol)

=> R là Cu

Câu 12: D

\(n_C=\dfrac{400.90\%}{12}=30\left(mol\right)\)

PTHH: C + O₂ --t^o--> CO₂

           30->30

=> V_O2 = 30.22,4 = 672 (l)