Stoichiometry Calculation Oxygen And Hydrogen Reaction Problem

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Hey guys! Today, we're diving into a classic stoichiometry problem involving the reaction between oxygen and hydrogen. This is a fundamental concept in chemistry, and mastering it will help you tackle more complex problems down the road. Let's break down the problem step by step to make sure we understand it completely. We'll be working through a specific example, so grab your calculators and let's get started!

Problem Statement

We're given that 16 grams of oxygen (O₂) reacts with 12 grams of hydrogen (H₂). Our mission is to figure out a few key things:

  1. Which reactant, oxygen or hydrogen, is in excess? In other words, which one do we have more of than we need for the reaction to go to completion?
  2. What is the amount of substance (in moles) and mass (in grams) of the excess reactant? This will tell us exactly how much of the excess reactant is left over after the reaction.
  3. What is the mass of water (H₂O) formed in the reaction? This is the product of the reaction, and we want to know how much we'll get.

1. Balanced Chemical Equation

The very first thing we need to do in any stoichiometry problem is write out the balanced chemical equation for the reaction. This tells us the exact ratio in which the reactants combine and the products are formed. For the reaction between hydrogen and oxygen to form water, the balanced equation is:

2 H₂ + O₂ → 2 H₂O

This equation tells us that 2 moles of hydrogen gas (H₂) react with 1 mole of oxygen gas (O₂) to produce 2 moles of water (H₂O). The coefficients in front of each chemical formula are crucial for determining the stoichiometry of the reaction.

2. Convert Mass to Moles

Since the balanced equation deals with moles, we need to convert the given masses of oxygen and hydrogen into moles. To do this, we'll use the molar masses of each substance.

  • The molar mass of hydrogen (H₂) is approximately 2 g/mol.
  • The molar mass of oxygen (O₂) is approximately 32 g/mol.

Now, let's calculate the number of moles:

  • Moles of H₂ = mass of H₂ / molar mass of H₂ = 12 g / 2 g/mol = 6 moles
  • Moles of O₂ = mass of O₂ / molar mass of O₂ = 16 g / 32 g/mol = 0.5 moles

So, we have 6 moles of hydrogen and 0.5 moles of oxygen.

3. Determine the Limiting Reactant

The limiting reactant is the reactant that gets used up first in the reaction. It determines the maximum amount of product that can be formed. To find the limiting reactant, we compare the mole ratio of the reactants to the stoichiometric ratio from the balanced equation.

From the balanced equation (2 H₂ + O₂ → 2 H₂O), the stoichiometric ratio of H₂ to O₂ is 2:1. This means that for every 1 mole of O₂, we need 2 moles of H₂.

Now, let's see if our amounts match this ratio:

  • We have 0.5 moles of O₂. According to the stoichiometry, we would need 2 * 0.5 = 1 mole of H₂ to react completely with all the oxygen.
  • We actually have 6 moles of H₂, which is way more than the 1 mole we need.

Since we have more hydrogen than we need, oxygen is the limiting reactant. This means that the reaction will stop when all the oxygen is used up.

4. Identify the Excess Reactant

The excess reactant is the reactant that is present in more than the required amount for the reaction. In this case, since oxygen is the limiting reactant, hydrogen is the excess reactant.

5. Calculate the Amount and Mass of Excess Reactant

To calculate the amount of excess reactant, we first need to determine how much hydrogen actually reacted with the 0.5 moles of oxygen. As we determined earlier, 0.5 moles of O₂ will react with 1 mole of H₂.

  • Moles of H₂ reacted = 1 mole

Now, we can calculate the moles of H₂ in excess:

  • Moles of H₂ in excess = initial moles of H₂ - moles of H₂ reacted = 6 moles - 1 mole = 5 moles

So, we have 5 moles of hydrogen in excess. To find the mass of this excess hydrogen, we multiply the moles by the molar mass:

  • Mass of H₂ in excess = moles of H₂ in excess * molar mass of H₂ = 5 moles * 2 g/mol = 10 grams

Therefore, there are 5 moles or 10 grams of hydrogen in excess.

6. Calculate the Mass of Water Formed

Now that we know the limiting reactant (oxygen), we can calculate the mass of water formed. From the balanced equation, 1 mole of O₂ produces 2 moles of H₂O. Since we have 0.5 moles of O₂, we will produce:

  • Moles of H₂O formed = 2 * moles of O₂ = 2 * 0.5 moles = 1 mole

The molar mass of water (H₂O) is approximately 18 g/mol. So, the mass of water formed is:

  • Mass of H₂O formed = moles of H₂O formed * molar mass of H₂O = 1 mole * 18 g/mol = 18 grams

Therefore, 18 grams of water are formed in this reaction.

Final Answers

Let's summarize our findings:

  1. Excess Reactant: Hydrogen (H₂)
  2. Amount and Mass of Excess Reactant: 5 moles or 10 grams of H₂
  3. Mass of Water Formed: 18 grams of H₂O

Key Concepts and Takeaways

Stoichiometry problems might seem daunting at first, but breaking them down into smaller steps makes them much more manageable. Here are the key concepts we used in this problem:

  • Balanced Chemical Equation: This is the foundation of all stoichiometry calculations. Make sure your equation is balanced before proceeding.
  • Moles: Moles are the central unit in stoichiometry. Convert masses to moles to use the stoichiometric ratios from the balanced equation.
  • Limiting Reactant: The limiting reactant determines the maximum amount of product that can be formed. It's essential to identify it correctly.
  • Excess Reactant: The reactant that is present in more than the required amount. Knowing the excess reactant helps you calculate how much is left over.

Pro Tips for Stoichiometry Success

  • Always start with a balanced equation: A correct balanced equation is crucial for accurate calculations.
  • Pay attention to units: Make sure your units are consistent throughout the problem. Convert masses to moles when necessary.
  • Show your work: Writing down each step helps you keep track of your calculations and makes it easier to spot mistakes.
  • Practice, practice, practice: The more stoichiometry problems you solve, the better you'll become at it.

Let's Chat!

Do you guys have any questions about this problem? Or maybe you've encountered a similar stoichiometry problem that you'd like to discuss? Drop a comment below, and let's learn together! Chemistry can be challenging, but it's also super rewarding when you start to understand how things work at the molecular level. Keep up the great work, and I'll see you in the next explanation! 📝✨