Respuesta :
Answer : The limiting reagent is [tex]O_2[/tex]
Solution : Given,
Moles of methane = 2.8 moles
Moles of [tex]O_2[/tex] = 5 moles
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]O_2[/tex] react with 1 mole of [tex]CH_4[/tex]
So, 5 moles of [tex]O_2[/tex] react with [tex]\frac{5}{2}=2.5[/tex] moles of [tex]CH_4[/tex]
From this we conclude that, [tex]CH_4[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/tex] is a limiting reagent and it limits the formation of product.
Hence, the limiting reagent is [tex]O_2[/tex]
