Question

Carbon monoxide, \(\mathrm{CO}\), and hydrogen react according to the equation below. $$3 \mathrm{CO}(\mathrm{g})+7 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ What volume of which reactant gas remains if \(12.0 \mathrm{LCO}(\mathrm{g})\) and \(25.0 \mathrm{L} \mathrm{H}_{2}(\mathrm{g})\) are allowed to react? Assume that the volumes of both gases are measured at the same temperature and pressure.

Short answer

The limiting reactant is H2 and after the reaction, 1.3 L of CO remains.

Step-by-step solution

Identify the Mole Ratio

The balanced equation tells us that 3 moles of CO react with 7 moles of H2 to give products. So, the mole ratio of CO to H2 is 3:7.

Calculate Reactant Ratios

According to the given exercise, we have 12.0 L of CO and 25.0 L of H2. Considering the mole ratio from step 1, we can say that for every 3 L of CO we need 7 L of H2.

Determine the Limiting Reactant

To find the limiting reactant, compare the ratio of reactants provided with the ratio required by the equation. In this case, (25.0L H2)/(12.0L CO) = 2.08 and (7 moles H2)/(3 moles CO) = 2.33. Since the given ratio is less than the required ratio, we know that H2 is the limiting reactant, and CO is in excess.

Calculate the Remaining Volume of CO

Since we know that H2 is the limiting reactant and 25.0 L of H2 react completely, according to the ratio in the balanced equation, we would need (3 moles CO/ 7 moles H2) x 25.0 L H2 = 10.7 L CO. Initially, we had 12.0 L of CO, so the remaining CO would be 12.0 L - 10.7 L = 1.3 L.

Key concepts

Limiting Reactant

In chemistry, the limiting reactant is the substance that gets entirely consumed first in a chemical reaction. This reactant limits the amount of products formed. Once the limiting reactant is used up, the reaction stops, even if other reactants are still available.
The concept of the limiting reactant is crucial for predicting how much product will be produced in a reaction. It helps to ensure that reactions are efficient and resources are not wasted.
To identify the limiting reactant, you need to compare the actual mole ratio of the reactants with the required mole ratio from the balanced equation. In the provided exercise, the given ratio of \[\frac{25.0\, \text{L}\, \text{H}_2}{12.0\, \text{L}\, \text{CO}} = 2.08\]is less than the required ratio \[\frac{7\, \text{mol}\, \text{H}_2}{3\, \text{mol}\, \text{CO}} = 2.33.\]This indicates that hydrogen (\(\text{H}_2\)) is the limiting reactant. Once \(\text{H}_2\) is used up, the reaction can no longer proceed and some \(\text{CO}\) will remain.

Balanced Chemical Equation

A balanced chemical equation represents a chemical reaction using symbols and formulas. It shows the reactants transforming into products, with both sides having an equal number of each type of atom.
This balance is necessary to satisfy the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction. To balance an equation, adjust coefficients to have the same number of each type of atom on both sides.
This process involves:

• Identifying each compound's elements in the reaction.
• Counting the atoms of each element on both sides of the equation.
• Adjusting coefficients to make these counts equal.

In the given exercise, the equation \(3\, \text{CO} + 7\, \text{H}_2 \rightarrow \text{C}_3\text{H}_8 + 3\, \text{H}_2\text{O}\)is balanced, meaning it correctly represents the conservation of mass.

Mole Ratio

The mole ratio in a chemical equation is the ratio between the amounts in moles of any two compounds involved in a chemical reaction. It is derived from the coefficients of a balanced chemical equation.
The mole ratio is essential for calculating how much of one reactant is required to react with a given amount of another reactant, or to predict the amount of product formed.
In the exercise, the mole ratio from the balanced equation is:\[3\, \text{mol}\, \text{CO} : 7\, \text{mol}\, \text{H}_2.\]This ratio tells us how many moles of \(\text{H}_2\) are needed for every mole of \(\text{CO}\), and vice versa. By comparing this with the actual ratio of reactants provided, we can determine the limiting reactant and how much of the non-limiting reactant remains.