Question

Methane, CH 4 , burns in oxygen to give carbon dioxide and water according to the following equation:

CH 4 + 2 O 2 ------> CO 2 + 2 H 2 O
In one experiment, a mixture of 0.250 mol of methane was burned in 1.25 mol of oxygen in a sealed steel
vessel. Find the limiting reactant and excess reactants.

Answer 1

Answer:
`CH_4` is the limiting reagent and
`O_2` is the excess reagent.

Step-by-step explanation:

The balanced chemical reaction is :

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

According to stoichiometry :

1 mole of
`CH_4` require = 2 moles of
`O_2`

Thus 0.250 moles of
`CH_4` will require=
`(2)/(1)* 0.250=0.500moles` of
`O_2`

As
`O_2` present is more than the required amount , it is an excess reagent.

Thus
`CH_4` is the limiting reagent as it limits the formation of product and
`O_2` is the excess reagent.