Question

What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?

Answer 1

Answer : The excess reactant in the combustion of methane in opem atmosphere is
`O_(2)` 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,

`CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)`

First we have to calculate the moles of methane.

`\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}}` =
`(23g)/(16.04g/mole)` = 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
`(2moles* 1.434moles)/(1moles)` 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
`O_(2)` molecule.