Understanding the Chemical Reaction

The given chemical equation represents the reaction between methane (\(C{H_4}\)) and oxygen (\({O_2}\)) to produce carbon monoxide (\(CO\)) and hydrogen (\({H_2}\)).

The balanced equation is:

\(C{H_4}\left( g \right) + \frac{1}{2}{O_2} \to CO\left( g \right) + 2{H_2}\left( g \right)\)

This equation tells us the mole ratio of reactants and products involved in the reaction. From the stoichiometry:

• 1 mole of \(C{H_4}\) reacts with 0.5 moles of \({O_2}\).
• 1 mole of \(C{H_4}\) produces 1 mole of \(CO\).
• 0.5 moles of \({O_2}\) produces 1 mole of \(CO\).

Identifying the Limiting Reactant

We are given specific initial amounts of the reactants: 2.0 moles of \(C{H_4}\) and 2.0 moles of \({O_2}\). In reactions where specific amounts of reactants are mixed, one reactant might run out before the other. The reactant that gets completely consumed first is called the limiting reactant. The limiting reactant determines the maximum amount of product that can be formed.

To find the limiting reactant, we can compare the ratio of the initial moles of each reactant to their stoichiometric coefficients in the balanced equation. The reactant with the smallest ratio is the limiting reactant.

• For \(C{H_4}\): Initial moles = 2.0 mol, Stoichiometric coefficient = 1
• Ratio for \(C{H_4}\): \(\frac{2.0 \text{ mol}}{1} = 2.0\)

• For \({O_2}\): Initial moles = 2.0 mol, Stoichiometric coefficient = \(\frac{1}{2} = 0.5\)
• Ratio for \({O_2}\): \(\frac{2.0 \text{ mol}}{0.5} = 4.0\)

Comparing the ratios, \(2.0\) (for \(C{H_4}\)) is less than \(4.0\) (for \({O_2}\)). Therefore, \(C{H_4}\) is the limiting reactant.

Alternatively, we can determine how much of one reactant is needed to react completely with the other. Let's see how much \({O_2}\) is needed to react with 2.0 moles of \(C{H_4}\):

From the equation, 1 mole of \(C{H_4}\) reacts with 0.5 moles of \({O_2}\). So, 2.0 moles of \(C{H_4}\) will react with \(2.0 \text{ mol } C{H_4} \times \frac{0.5 \text{ mol } O_2}{1 \text{ mol } C{H_4}} = 1.0 \text{ mol } O_2\).

We have 2.0 moles of \({O_2}\) available, which is more than the 1.0 mole needed. This confirms that \({O_2}\) is in excess and \(C{H_4}\) is the limiting reactant.

Calculating the Moles of CO Produced

Since \(C{H_4}\) is the limiting reactant, the amount of \(CO\) produced depends on the initial amount of \(C{H_4}\).

From the balanced equation, 1 mole of \(C{H_4}\) produces 1 mole of \(CO\). Therefore, 2.0 moles of \(C{H_4}\) will produce:

\(2.0 \text{ mol } C{H_4} \times \frac{1 \text{ mol } CO}{1 \text{ mol } C{H_4}} = 2.0 \text{ mol } CO\)

The maximum number of moles of \(CO\) that can be obtained is 2.0 moles.

Conclusion

Based on the limiting reactant analysis using the given amounts of 2.0 moles of \(C{H_4}\) and 2.0 moles of \({O_2}\), the maximum number of moles of \(CO\) that can be produced according to the reaction \(C{H_4}\left( g \right) + \frac{1}{2}{O_2} \to CO + 2{H_2}\) is 2.0 moles.

Revision Table: Stoichiometry and Limiting Reactant

Additional Information: Calculating Product Yield

The amount of product calculated based on the limiting reactant is the theoretical yield – the maximum possible amount of product that can be formed under ideal conditions. In a real experiment, the actual yield (the amount of product isolated) is often less than the theoretical yield due to various factors like incomplete reactions, side reactions, or loss during purification.

The percent yield is a measure of the efficiency of a reaction, calculated as:

Percent Yield = \(\left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%\)

Understanding stoichiometry and limiting reactants is crucial for predicting reaction outcomes and optimizing chemical processes.