Final answer:
To find the mass of methane left over, the masses of methane and oxygen must be converted to moles to identify the limiting reactant. As methane is the limiting reactant and will be completely consumed in the reaction, there will be no methane left over. Therefore, the minimum mass of methane that could be left over by the chemical reaction is 0 g.
Step-by-step explanation:
The student is asking how to calculate the minimum mass of methane that could be left over after it reacts with oxygen to produce carbon dioxide and water. To solve this problem, we first need to consider the balanced chemical equation for the reaction:
CH4 + 2 O2 → CO2 + 2 H2O
We'll need to convert the masses of methane and oxygen to moles to see which reactant is the limiting reactant. The molar mass of methane (CH4) is approximately 16.04 g/mol, and the molar mass of oxygen (O2) is approximately 32.00 g/mol. Therefore:
- 8.66 g CH4 ÷ 16.04 g/mol = 0.54 moles of CH4
- 57.0 g O2 ÷ 32.00 g/mol = 1.78 moles of O2
According to the stoichiometry of the reaction, 1 mole of methane reacts with 2 moles of oxygen. Thus, 0.54 moles of methane would require 1.08 moles of oxygen to react completely. Since we have 1.78 moles of oxygen available, oxygen is in excess, and methane is the limiting reactant.
Since all the methane will be consumed, there will be no methane left over, and the minimum mass of methane that could be left over is 0 g.