Question

If 16.0 g of methane is allowed to react with 32.0 g of oxygen, what is the limiting reactant? Explain.

{CH}_4(g) + 2{O}_2(g) → {CO}_2 (g) + 2{H}_2{O}(g)

A. Methane is the limiting reactant because it has less mass.
B. Oxygen is the limiting reactant because it has more mass.
C. Methane is the limiting reactant because it has a lower molar mass.
D. Oxygen is the limiting reactant because it has a higher molar mass

Answer 1

To identify the limiting reactant between methane and oxygen, stoichiometric calculations based on their given masses and the balanced equation are used. Oxygen is the limiting reactant because there is not enough of it to fully react with the available methane.

Step-by-step explanation:

In order to find the limiting reactant, we need to perform stoichiometric calculations based on the balanced chemical equation CH4(g) + 2O2(g) → CO2 (g) + 2H2O(g). Starting with 16.0 g of methane (CH4), and using its molar mass (16.04 g/mol), we find that there are 1 mol of CH4 present.

Similarly, starting with 32.0 g of oxygen (O2), using the molar mass of O2 (32.00 g/mol), we calculate there are 1 mol of O2. Since the stoichiometry requires 2 moles of O2 for every mole of CH4, oxygen is the limiting reactant as we would have needed 32.08 g of O2 to completely react with the methane.