2 C₂H₆ + 5 O₂ → 4 CO + 6 H₂O\
[tex]M_{C2H6}[/tex] = 2 · 12g/mol + 6 · 1g/mol = 30g/mol
[tex]M_{CO} =[/tex] 12g/mol + 16g/mol = 28g/mol
60g C₂H₆ ------ x g CO
2 · 30 g ------ 4 · 28 g
x= [tex]\frac{60 * (4 * 28)}{2 * 30} = \frac{60*112}{60}=\frac{6720}{60}=112 g[/tex]
odp. W wyniku spalania niecałkowitego powstanie 112 g tlenku węgla