When 2.05 moles of CH4 are mixed with 5.03 moles of O2 in the reaction below, the limiting reactant is
CH4 + 2O2 -> CO2 + 2H2O
To determine the limiting reactant in a chemical reaction, we need to compare the stoichiometric ratio of the reactants to the given amounts.
In the given reaction: CH4 + 2O2 -> CO2 + 2H2O
The stoichiometric ratio between CH4 and O2 is 1:2. This means that for every 1 mole of CH4, we need 2 moles of O2 to react completely.
Given:
Moles of CH4 = 2.05 moles
Moles of O2 = 5.03 moles
To find out which reactant is limiting, we need to calculate the number of moles of each reactant required based on the stoichiometric ratio.
Moles of O2 required = 2 * Moles of CH4 = 2 * 2.05 moles = 4.10 moles
Since we have 5.03 moles of O2, which is more than the required amount, O2 is in excess.
Therefore, the limiting reactant is CH4.
To determine the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation.
In the balanced equation, the stoichiometric coefficient in front of CH4 is 1, which means that 1 mole of CH4 reacts with 2 moles of O2.
Now let's calculate the number of moles of O2 required to fully react with 2.05 moles of CH4. Since the ratio of moles between CH4 and O2 is 1:2, we can multiply the moles of CH4 by 2 to find the moles of O2 required:
2.05 moles CH4 * 2 moles O2/1 mole CH4 = 4.10 moles O2
Next, we compare this value to the actual amount of O2 present, which is 5.03 moles.
Since 5.03 moles of O2 is greater than 4.10 moles O2, the amount of O2 is in excess. Therefore, CH4 is the limiting reactant.
CH4 is the limiting reactant
I agree that CH4 is the LR but here is how you prove it.
Convert 2.05 mols CH4 to mols needed for O2.
2.05 mols CH4 x (2 mols O2/1 mol CH4) = 4.10 mols O2 needed. Do you have that much? Yes. So CH4 is the limiting reagent.