if 40.0 g of methane gas reacts with 40.0 g of oxygen gas whis it the limiting reactant

1. Write and balance the equation.

2a. mols O2 = grams/molar mass
2b. mols CH4 = grams/molar mass.
3a. Convert mols O2 to mols of a product.
3b. Do the same for mols CH4.
3c. The limiting reagent in these problems is ALWAYS the reagent that provides the FEWER mols product.

To determine the limiting reactant in a chemical reaction, follow these steps:

1. Write and balance the chemical equation for the reaction. In this case, the balanced equation for the reaction between methane (CH4) and oxygen (O2) to form carbon dioxide (CO2) and water (H2O) is:
CH4 + 2O2 → CO2 + 2H2O

2. Calculate the molar mass of each reactant. The molar mass of methane (CH4) is 16.04 g/mol, and the molar mass of oxygen (O2) is 32.00 g/mol.

3. Convert the mass of each reactant to moles using the formula:
moles = mass (g) / molar mass (g/mol)

For methane:
moles of CH4 = 40.0 g / 16.04 g/mol ≈ 2.49 mol

For oxygen:
moles of O2 = 40.0 g / 32.00 g/mol ≈ 1.25 mol

4. Examine the stoichiometry of the balanced equation to determine the ratio of reactants needed for the reaction. In this case, the ratio between methane and oxygen is 1:2. This means that 1 mole of methane reacts with 2 moles of oxygen.

5. Compare the actual ratio of moles of reactants to the ratio determined by stoichiometry. In this case, we have 2.49 moles of CH4 and 1.25 moles of O2.
The ratio of CH4 to O2 is:
2.49 mol CH4 / 1.25 mol O2 ≈ 1.992

6. The reactant with the smaller ratio is the limiting reactant, as it will be completely consumed first and will determine the maximum amount of product that can be formed. In this case, oxygen has a smaller ratio of 1.25, indicating that it is the limiting reactant. Methane is in excess.

Therefore, the limiting reactant in this reaction is oxygen gas.