To calculate the minimum mass of methane that could be left over after the chemical reaction, we need to determine which reactant is limiting, which means it will be completely consumed in the reaction. We can do this by comparing the stoichiometry of the reaction to the amounts of reactants given.
The balanced chemical equation for the reaction between methane (CH4) and oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is:
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that one mole of methane (CH4) reacts with two moles of oxygen (O2). We need to calculate the number of moles for each reactant:
1. Moles of CH4:
Mass of CH4 = 3.7 g
Molar mass of CH4 = 12.01 g/mol (C) + 4.02 g/mol (H) = 16.03 g/mol
Moles of CH4 = 3.7 g / 16.03 g/mol ≈ 0.231 mol
2. Moles of O2:
Mass of O2 = 22.2 g
Molar mass of O2 = 2 * 16.00 g/mol = 32.00 g/mol
Moles of O2 = 22.2 g / 32.00 g/mol ≈ 0.694 mol
Now, we need to find the mole ratio between CH4 and O2 from the balanced equation. It's 1:2, which means one mole of CH4 reacts with two moles of O2. However, we have less than 0.231 moles of CH4 for the reaction.
To react completely, 0.231 moles of CH4 would require (0.231 mol CH4) * (2 mol O2/mol CH4) = 0.462 moles of O2. Since we have only 0.694 moles of O2, which is more than enough to react with the available CH4, CH4 is the limiting reactant.
Now, we can calculate the mass of CH4 that reacts completely:
Moles of CH4 * Molar mass of CH4 = 0.231 mol * 16.03 g/mol = 3.70 g (rounded to two significant digits)
So, all of the 3.7 g of methane will react, and none will be left over.