Question

Answer 1

\(n_{NO} = \dfrac{0,896}{22,4} =0,04(mol)\\ N^{+5} + 3e \to N^{+2}\)

\(n_{NO_3^-} = n_e = 3n_{NO} = 0,04.3 = 0,12(mol)\)

\(\Rightarrow m_{muối} = m_{kim\ loại} + m_{NO_3^-} = 2,06 + 0,12.62 = 9,5(gam)\)

(Đáp án A)

Answer 2

\(N^{+5}+3e\rightarrow N^{+2}\)

............0,12..0,04

\(\Rightarrow m_M=m_{KL}+m_{NO3}=2,06+7,44=9,5\left(g\right)\)

=> Đáp án A.