During a combustion reaction, 5.00 grams of oxygen reacted with 5.00 grams of ch4.
what is the amount of the leftover reactant?
4.00 grams of methane
3.75 grams of methane
2.25 grams of oxygen
1.75 grams of oxygen

3.75 grams of methane

Explanation:

Given parameters:

Mass of oxygen gas = 5g

Mass of CH₄ = 5g

Unknown parameters:

Amount of excess reactant = ?

Solution

In order to find the amount of leftover reactant we need to know  the reactant that is in excess and in limited supply. The amount of reactant in short supply will determine the extent of the reaction.

Now, we need to write the balanced reaction equation:

                   CH₄ + 2O₂ → CO₂ + 2H₂O

We first find the number of moles of the reactants.

Number of moles of methane =

Number of moles of methane =   = 0.3125mole

Number of moles of oxygen = = 0.156mole

    From the equation:

      1 mole of methane reacts with 2 moles of oxygen gas

      x moles of methane reacts with 0.156moles of oxygen gas

                2x = 0.156

                  x = 0.078moles

We see that the methane is in excess and oxygen gas is in short supply in the reaction.

The amount of excess methane = 0.3125 - 0.078 = 0.2345moles

Now we convert it to mass:

  Mass = number of moles x molar mass = 0.2345 x 16 = 3.75grams