$\rm n_{Fe}=\dfrac{0,84}{56}=0,015\ (mol)\\n_{Cu}=\dfrac{1,92}{64}=0,03\ (mol)\\n_{NO_2}=\dfrac{1,008}{22,4}=0,045\ (mol)$

Quá trình nhường electron:

$\rm\mathop{Fe}\limits^{0}\to \mathop{Fe}\limits^{+3}+3e\\\mathop{Cu}\limits^{0}\to \mathop{Cu}\limits^{+2}+2e$

Quá trình nhận electron:

$\rm\mathop{N}\limits^{+5}+e\to \mathop{N}\limits^{+4}\\\mathop{N}\limits^{+5}+3e\to \mathop{N}\limits^{+2}$

Bảo toàn electron:

$\rm 3×0,015+2×0,03=0,045+3n_{NO}\to n_{NO}=0,02\ (mol)\\V_{NO}=0,02×22,4=0,448\ (l)$

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