Gọi \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) => 24a + 27b = 3,72 (*)
Quá trình oxi hoá - khử:
\(Mg^0\rightarrow Mg^{+2}+2e\\ Al^0\rightarrow Al^{+3}+3e\\ 2N^{+5}+8e\rightarrow2N^{+1}\)
Bảo toàn electron: \(2n_{Mg}+3n_{Al}=8n_{N_2O}\)
=> 2x + 3y = 0,4 (**)
Từ (*), (**) => x = 0,02; y = 0,12
BTNT Mg, Al: \(\left\{{}\begin{matrix}n_{Al\left(NO_3\right)_3}=n_{Al}=0,12\left(mol\right)\\n_{Mg\left(NO_3\right)_2}=n_{Mg}=0,02\left(mol\right)\end{matrix}\right.\)
BTNT N: \(n_{NO_2}=3n_{Al\left(NO_3\right)_3}+2n_{Mg\left(NO_3\right)_2}=0,4\left(mol\right)\)
Quá trình oxi hoá - khử:
\(N^{+5}+1e\rightarrow N^{+4}\\ 2O^{-2}\rightarrow O_2\uparrow+4e\)
BTe: \(n_{O_2}=\dfrac{1}{4}n_{NO_2}=0,1\left(mol\right)\)
=> \(M_C=\dfrac{0,1.32+0,4.46}{0,1+0,4}=43,2\left(g/mol\right)\)
=> \(d_{C/H_2}=\dfrac{43,2}{2}=21,6\)
Chọn A
$n_{N_2O} = \dfrac{1120}{1000.22,4} = 0,05(mol)$
Ta có :
$n_{NO_3\ trong\ muối} = n_{electron\ trao\ đổi} = 8n_{N_2O} = 0,4(mol)$
$4NO_3 \to 2O_{oxit} + 4NO_2 + O_2$
Ta có :
$n_{NO_2} = n_{NO_3} = 0,4(mol)$
$n_{O_2} = \dfrac{1}{4}n_{NO_3} = 0,1(mol)$
$M_C = \dfrac{0,4.46 + 0,1.32}{0,4 + 0,1} = 43,2(g/mol)$
$d_{C/H_2} = \dfrac{43,2}{2} = 21,6$
