Question

13) Methane (CH4) combines with oxygen according to the balanced equation below. What is the limiting reactant if 25.0 g of CH4 combines with 50.0 g of O2? CH4 + 2 O2 → CO2 + 2 H2O moles CH4 moles O2 moles O2 needed to react with given amount of CH4 limiting reactant ("methane" or "oxygen")

Answer 1

Limiting reactant: Oxygen
`O_2`

Step-by-step explanation:

First write the balanced chemical equation:

`CH_4 + 2 O_2 \rightarrow CO_2 + 2 H_2O`

`mole=(given\, mass)/(molecular\, mass)`

given mass of methane =25g

molecular mass of methane=16g/mol

mole=1.5625mol

given mass of oxygen=50

molar mass of oxygen=32g/mol

mole=1.5625mol

from the above balanced equation it is clearly that,

1 mole of methane needs 2 moles of oxygen for complete reaction

Therefore,

1.5625 moles of methane needs 3.125 moles of oxygen for complete reaction but we have only 1.5625 moles of oxygen,

hence,

oxygen will be the limiting reactant and methane will be the excess reactant