Potassium Nitrate Production: Stoichiometry Problem

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Hey guys! Let's dive into a stoichiometry problem where we'll figure out how much potassium nitrate (KNO3) is produced when potassium phosphate (K3PO4) reacts with excess aluminum nitrate (Al(NO3)3). This is a classic chemistry problem that involves balancing equations, understanding mole ratios, and applying stoichiometry principles. So, grab your calculators and let's get started!

The Reaction: K3PO4 + Al(NO3)3 → KNO3 + AlPO4

Before we jump into calculations, we need to make sure we have a balanced chemical equation. Balancing chemical equations is crucial in stoichiometry because it ensures that the number of atoms for each element is the same on both sides of the equation, adhering to the law of conservation of mass. The unbalanced equation we have is:

K3PO4 + Al(NO3)3 → KNO3 + AlPO4

Let's balance this step by step.

  1. Count the atoms:

    • Left side: 3 K, 1 P, 4 O, 1 Al, 3 N, 9 O (from nitrate)
    • Right side: 1 K, 1 P, 4 O, 1 Al, 1 N, 3 O (from nitrate)

  2. Balance potassium (K): We have 3 K on the left and 1 K on the right. So, we add a coefficient of 3 in front of KNO3:

    K3PO4 + Al(NO3)3 → 3 KNO3 + AlPO4

  3. Count atoms again:

    • Left side: 3 K, 1 P, 4 O, 1 Al, 3 N, 9 O
    • Right side: 3 K, 1 P, 4 O, 1 Al, 3 N, 9 O

Great! The equation is now balanced. Our balanced chemical equation is:

K3PO4 + Al(NO3)3 → 3 KNO3 + AlPO4

This balanced equation tells us the mole ratios of the reactants and products. For every 1 mole of K3PO4 that reacts, we produce 3 moles of KNO3. This mole ratio is the key to solving our problem.

Understanding Moles and Stoichiometry

So, what exactly are moles? A mole is a unit of measurement used in chemistry to express amounts of a chemical substance. Specifically, it's defined as the amount of any substance that contains as many entities (e.g., atoms, molecules, ions) as there are atoms in 12 grams of pure carbon-12. This number is known as Avogadro's number, which is approximately 6.022 x 10^23.

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It's like the recipe book for chemical reactions. By using the balanced chemical equation, we can predict the amount of reactants needed and the amount of products formed in a chemical reaction.

The coefficients in a balanced chemical equation represent the mole ratios of the substances involved. In our balanced equation:

K3PO4 + Al(NO3)3 → 3 KNO3 + AlPO4

The coefficients are 1, 1, 3, and 1. This means:

  • 1 mole of K3PO4 reacts with 1 mole of Al(NO3)3
  • To produce 3 moles of KNO3
  • And 1 mole of AlPO4

These ratios are our conversion factors for calculating the amount of product formed from a given amount of reactant.

Calculating Moles of KNO3 Produced

We are given that 2.20 moles of K3PO4 react with excess Al(NO3)3. The term "excess" is important because it tells us that K3PO4 is the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, determining the maximum amount of product that can be formed. Since Al(NO3)3 is in excess, we have more than enough of it, so K3PO4 will run out first, and the amount of KNO3 produced will depend solely on the amount of K3PO4.

From the balanced equation, we know that 1 mole of K3PO4 produces 3 moles of KNO3. We can write this as a mole ratio:

(3 moles KNO3) / (1 mole K3PO4)

Now, we can use this ratio to find out how many moles of KNO3 are produced from 2.20 moles of K3PO4:

Moles of KNO3 = 2.20 moles K3PO4 * (3 moles KNO3 / 1 mole K3PO4) Moles of KNO3 = 2.20 * 3 moles Moles of KNO3 = 6.60 moles

So, 2.20 moles of K3PO4 will produce 6.60 moles of KNO3.

Converting Moles of KNO3 to Grams

We've calculated the moles of KNO3 produced, but the question asks for the mass in grams. To convert moles to grams, we need the molar mass of KNO3. The molar mass is the mass of one mole of a substance, and it's numerically equal to the substance's atomic or molecular weight in atomic mass units (amu). We can find the molar mass by adding up the atomic masses of each element in the compound from the periodic table.

KNO3 consists of:

  • 1 potassium (K): approximately 39.10 g/mol
  • 1 nitrogen (N): approximately 14.01 g/mol
  • 3 oxygen (O): approximately 3 * 16.00 = 48.00 g/mol

So, the molar mass of KNO3 is:

Molar mass of KNO3 = 39.10 g/mol + 14.01 g/mol + 48.00 g/mol Molar mass of KNO3 = 101.11 g/mol

Now, we can use the molar mass to convert moles of KNO3 to grams:

Mass of KNO3 = Moles of KNO3 * Molar mass of KNO3 Mass of KNO3 = 6.60 moles * 101.11 g/mol Mass of KNO3 = 667.326 g

Rounding this to three significant figures (since our initial value, 2.20 moles, has three significant figures), we get 667 g.

Final Answer

Therefore, when 2.20 moles of potassium phosphate react with excess aluminum nitrate, approximately 667 grams of potassium nitrate are produced. So the correct answer is B. 667 g.

Key Takeaways from this Stoichiometry Problem

  • Balancing Chemical Equations: Always make sure your equation is balanced before doing any stoichiometric calculations. The coefficients are the foundation of mole ratios.
  • Understanding Mole Ratios: The balanced equation tells you the mole ratios between reactants and products. Use these ratios to convert between moles of different substances.
  • Identifying the Limiting Reactant: If you're given amounts of multiple reactants, determine which one is the limiting reactant. This reactant determines the maximum amount of product that can be formed.
  • Converting Moles to Grams (and Vice Versa): Use the molar mass to convert between moles and grams. This is a fundamental skill in stoichiometry.

Stoichiometry might seem daunting at first, but with practice, it becomes second nature. Remember to break down the problem into steps, and you'll be solving complex chemistry problems in no time! Keep practicing, guys, and you'll become stoichiometry pros!