Respuesta :
Answer : The excess reactant in the combustion of methane in opem atmosphere is [tex]O_{2}[/tex] molecule.
Solution : Given,
Mass of methane = 23 g
Molar mass of methane = 16.04 g/mole
The Net balanced chemical reaction for combustion of methane is,
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)[/tex]
First we have to calculate the moles of methane.
[tex]\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}}[/tex] = [tex]\frac{23g}{16.04g/mole}[/tex] = 1.434 moles
From the above chemical reaction, we conclude that
1 mole of methane react with the 2 moles of oxygen
and 1.434 moles of methane react to give [tex]\frac{2moles\times 1.434moles}{1moles}[/tex] moles of oxygen
The Moles of oxygen = 2.868 moles
Now we conclude that the moles of oxygen are more than the moles of methane.
Therefore, the excess reactant in the combustion of methane in open atmosphere is [tex]O_{2}[/tex] molecule.
