\(n_X = \dfrac{0,986.1}{0,082.(273+27,3)} = 0,04(mol)\)
Gọi : \(n_{NO} = a(mol) ; n_{N_2} = b(mol) \Rightarrow a + b = 0,04(1)\)
\(m_X = 30a + 28b = 0,04.14,75.2=1,18(gam)\)
Suy ra a = 0,03 ; b = 0,01
Gọi : \(n_{Fe} = x(mol) ; n_{Mg} = y(mol)\)
Ta có :
\(56x + 24y = 2,88\\
\text{Bảo toàn e} : 3x + 2y = 3a + 10b = 0,19\\
\Rightarrow x = 0,03; y = 0,05\\
\%m_{Fe} = \dfrac{0,03.56}{2,88}.100\% = 58,33\%\\
\%m_{Mg} = 100\% -58,33\% = 41,67\%\)