In the reaction below, 25.0g of potassium superoxide, KO2, (molar mass is 71.10g/mole) is used with 45.0g of CO2 (molar mass is 44.01g/mole). how many grams of the excess reagent are left unreacted?
The Correct Answer and Explanation is:
To solve this problem, we need to analyze the chemical reaction between potassium superoxide (KO₂) and carbon dioxide (CO₂). The balanced chemical equation is:
$$
4KO_2 + 2CO_2 \rightarrow 2K_2CO_3 + 3O_2
$$
Step 1: Calculate the moles of each reactant
Moles of KO₂:
$$
\frac{25.0 \text{ g}}{71.10 \text{ g/mol}} = 0.3517 \text{ mol KO₂}
$$
Moles of CO₂:
$$
\frac{45.0 \text{ g}}{44.01 \text{ g/mol}} = 1.0227 \text{ mol CO₂}
$$
Step 2: Use the mole ratio from the balanced equation
From the equation:
$$
4 \text{ mol KO₂} : 2 \text{ mol CO₂}
$$
This simplifies to:
$$
2 \text{ mol KO₂} : 1 \text{ mol CO₂}
$$
So, for every 2 moles of KO₂, we need 1 mole of CO₂.
Step 3: Determine the limiting reagent
Using the mole ratio, check how much CO₂ is needed for the given KO₂:
$$
\frac{0.3517 \text{ mol KO₂}}{2} = 0.1759 \text{ mol CO₂ needed}
$$
We have 1.0227 mol CO₂ available, which is much more than 0.1759 mol, so KO₂ is the limiting reagent, and CO₂ is in excess.
Step 4: Calculate how much CO₂ is used
From the balanced reaction:
$$
4 \text{ mol KO₂} : 2 \text{ mol CO₂}
\Rightarrow \frac{0.3517 \text{ mol KO₂}}{4} \times 2 = 0.1759 \text{ mol CO₂ used}
$$
Step 5: Calculate how much CO₂ remains
$$
1.0227 \text{ mol CO₂} – 0.1759 \text{ mol CO₂} = 0.8468 \text{ mol CO₂ remaining}
$$
Convert to grams:
$$
0.8468 \text{ mol} \times 44.01 \text{ g/mol} = 37.25 \text{ g CO₂ unreacted}
$$
✅ Final Answer:
37.3 grams of CO₂ remain unreacted
In this problem, we are asked to determine how much of the excess reactant remains unreacted after the completion of a chemical reaction between potassium superoxide (KO₂) and carbon dioxide (CO₂). The balanced equation for the reaction is:
$$
4KO_2 + 2CO_2 \rightarrow 2K_2CO_3 + 3O_2
$$
The first step is to convert the masses of the two reactants into moles using their molar masses. Potassium superoxide has a molar mass of 71.10 g/mol, and carbon dioxide has a molar mass of 44.01 g/mol. Dividing the given masses by their respective molar masses gives us approximately 0.3517 moles of KO₂ and 1.0227 moles of CO₂.
Next, we use the mole ratio from the balanced equation to determine which reactant is limiting. The equation shows that 4 moles of KO₂ react with 2 moles of CO₂, or more simply, 2 moles of KO₂ for every 1 mole of CO₂. Based on this ratio, 0.3517 moles of KO₂ would need only about 0.1759 moles of CO₂ to react completely.
Since we have 1.0227 moles of CO₂—much more than needed—it is in excess, and KO₂ is the limiting reagent. This means all the KO₂ will be consumed, and not all the CO₂ will react.
We calculate the amount of CO₂ that actually reacts (0.1759 moles), subtract it from the original amount, and convert the leftover CO₂ back into grams. This yields approximately 37.3 grams of CO₂ remaining unreacted.
Understanding limiting and excess reagents is crucial in stoichiometry, as it allows us to predict the amounts of products and leftovers in a reaction.
