111k views
5 votes
If 16g of methane is allowed to react with 32g of oxygen what is the limiting factor

a) Identify the limiting reactant
b) Calculate the moles of CH 4
c) Calculate the moles of 02
d) Write the balanced chemical equation

User MaxSem
by
7.7k points

1 Answer

6 votes

Final answer:

The limiting reactant when 16g of methane reacts with 32g of oxygen is oxygen (O2). One mole of methane would require two moles of oxygen to completely react, but only one mole of oxygen is available, thus O2 is the limiting factor.

Step-by-step explanation:

Identifying the Limiting Reactant

To determine the limiting reactant, we must first write the balanced chemical equation for the reaction of methane (CH4) with oxygen (O2). Methane reacts with oxygen to produce carbon dioxide (CO2) and water (H2O):

CH4 + 2O2 → CO2 + 2H2O

Calculating Moles of Reactants

Determining the Limiting Reactant

From the balanced equation, it is clear that 1 mole of CH4 reacts with 2 moles of O2. Since we only have 1 mole of O2, the limiting reactant is O2 because there is not enough O2 to react with all the CH4 present.

User Chris Van Opstal
by
7.4k points