Select the correct answer.
If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate ($KNO_3$), what is the limiting reagent?
$2 KNO_3+10 K \rightarrow 6 K_2 O+N_2$
A. potassium nitrate $(KNO_3)$
B. potassium (K)
C. potassium oxide $(K_2 O)$
D. $(N_2)$
E. nitrate $(NO_3)$
Answer (1)
Calculate the moles of potassium (K): 39.10 g / m o l 155 g ≈ 3.96 m o l .
Calculate the moles of potassium nitrate ( K N O 3 ): 101.10 g / m o l 122 g ≈ 1.21 m o l .
Determine the required moles of K for the given moles of K N O 3 : 1.21 m o l × 5 = 6.05 m o l .
Since we only have 3.96 moles of K, which is less than the required 6.05 moles, potassium (K) is the limiting reagent. The answer is B .
Explanation
Problem Analysis We are given the reaction 2 K N O 3 + 10 K → 6 K 2 O + N 2 and the masses of the reactants, potassium (K) and potassium nitrate ( K N O 3 ). We need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed first and limits the amount of product formed. To find the limiting reagent, we need to compare the mole ratio of the reactants to the stoichiometric ratio from the balanced equation.
Calculating Moles of Reactants First, we calculate the number of moles of each reactant. The molar mass of potassium (K) is approximately 39.10 g/mol, and the molar mass of potassium nitrate ( K N O 3 ) is approximately 101.10 g/mol.
Moles of K = m o l a r ma ss ma ss = 39.10 g / m o l 155 g ≈ 3.96 m o l
Moles of K N O 3 = m o l a r ma ss ma ss = 101.10 g / m o l 122 g ≈ 1.21 m o l
Determining the Limiting Reagent From the balanced equation, the stoichiometric ratio of K to K N O 3 is 10:2, which simplifies to 5:1. This means that 5 moles of K are required to react with 1 mole of K N O 3 .
Now, we compare the actual mole ratio to the stoichiometric ratio. We have 3.96 moles of K and 1.21 moles of K N O 3 . If K N O 3 is the limiting reagent, we would need 1.21 m o l × 5 = 6.05 m o l of K. However, we only have 3.96 moles of K, which is less than the required 6.05 moles. Therefore, K is the limiting reagent.
Confirmation Alternatively, if K is the limiting reagent, we would need 5 3.96 m o l = 0.79 m o l of K N O 3 . We have 1.21 moles of K N O 3 , which is more than the required 0.79 moles. This confirms that K is the limiting reagent.
Final Answer Therefore, the limiting reagent is potassium (K). The correct answer is B.
Examples
In chemical manufacturing, determining the limiting reagent is crucial for optimizing production. For example, if you're producing fertilizers and one of the raw materials is more expensive or harder to obtain, you'd want to ensure it's the limiting reagent. This way, you minimize waste and maximize the use of the costly material, leading to cost-effective production. Understanding stoichiometry and limiting reagents helps in scaling up reactions efficiently and predicting yields accurately.
