Question

Answer 1

\(n_{Al}=x\left(mol\right),n_{Mg}=y\left(mol\right)\)

\(n_{H_2}=1.5x+y=0.2\left(mol\right)\)

\(BTe:\)

\(3x+2y=n\cdot n_X\)

\(\Rightarrow0.4=n\cdot0.05\)

\(\Rightarrow n=8\)

\(\Rightarrow B\)

Answer 2

\(2H^+ + 2e \to H_2\\ n_e = 2n_{H_2} = 2.\dfrac{4,48}{22,4} = 0,4(mol)\\ n_X = \dfrac{1,12}{22,4} = 0,05(mol)\)

Ta có :

\(\dfrac{n_e}{n_X} = \dfrac{0,4}{0,05} = 8\\ 2N^{+5} + 2e \to 2N^+\)

Vậy X là N₂O(Đáp án B)