How Many Grams Of Oxygen Gas Are Produced When 2.43

How Many Gramsof Oxygen Gas Are Produced When 2.43 g of a Reactant Decomposes? A Step‑by‑Step Guide to Stoichiometric Calculations

When faced with a question such as “how many grams of oxygen gas are produced when 2.43 g …?” the underlying task is a classic stoichiometry problem. You start with a known mass of a reactant, convert it to moles, use the balanced chemical equation to find the mole ratio to oxygen, and finally convert those moles of O₂ back into grams. This article walks you through the entire process, illustrates it with three common laboratory reactions, and highlights pitfalls to avoid—all in clear, easy‑to‑follow language.

Understanding the Core Concept: Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It rests on two pillars:

1. Conservation of mass – atoms are neither created nor destroyed.
2. Mole concept – one mole of any substance contains Avogadro’s number (≈ 6.022 × 10²³) of entities and has a mass equal to its molar mass (g mol⁻¹).

To answer “how many grams of oxygen gas are produced when 2.43 g …?” you follow these four logical steps:

Step‑by‑Step Calculation Method

Below is a generic template you can plug any reactant into. Replace the symbols with the actual formulas and numbers for your specific problem.

1. Write the balanced chemical equation for the reaction that produces O₂.
   Example:

2Mg(s) + O₂(g) → 2MgO(s)

2. Calculate the molar mass of the reactant.

   • Identify the chemical formula of the reactant.
   • Look up the atomic masses of each element in the formula on the periodic table.
   • Multiply each atomic mass by the number of atoms of that element in the formula.
   • Sum the results to get the molar mass of the reactant in g/mol.

3. Convert the given mass of the reactant to moles.

   • Divide the given mass of the reactant (in grams) by its molar mass (in g/mol).
   • This will give you the number of moles of the reactant.

4. Use the mole ratio from the balanced equation to find the moles of O₂ produced.

   • The balanced equation shows the stoichiometric relationship between the reactant and O₂.
   • Determine the mole ratio of reactant to O₂. For example, in the equation above, the mole ratio is 2 moles of Mg to 1 mole of O₂.

5. Convert the moles of O₂ to grams.

   • Multiply the moles of O₂ (calculated in step 4) by the molar mass of O₂ (32.00 g/mol).
   • This will give you the mass of O₂ produced in grams.

6. Check significant figures and units.

   • The final answer should have the same number of significant figures as the least precise measurement in the problem.
   • Ensure the units are correct (grams for mass).

Example 1: Decomposition of Magnesium

Let's calculate the grams of oxygen produced when 2.43 g of magnesium (Mg) decomposes.

1. Balanced Equation: 2Mg(s) + O₂(g) → 2MgO(s)
2. Molar Mass of Mg: Mg = 24.31 g/mol
3. Moles of Mg: 2.43 g Mg / 24.31 g/mol = 0.100 mol Mg
4. Mole Ratio: From the balanced equation, the mole ratio of Mg to O₂ is 2:1. Therefore, 0.100 mol Mg * (1 mol O₂ / 2 mol Mg) = 0.050 mol O₂
5. Grams of O₂: 0.050 mol O₂ * 32.00 g/mol = 1.60 g O₂
6. Significant Figures & Units: The given mass (2.43 g) has three significant figures, so the answer should also have three significant figures. The units are grams, which is correct.

Therefore, 1.60 grams of oxygen gas are produced when 2.43 g of magnesium decomposes.

Example 2: Decomposition of Potassium Chlorate

Consider the decomposition of potassium chlorate (KClO₃) to produce oxygen gas and potassium chloride (KCl):

2KClO₃(s) → 2KCl(s) + 3O₂(g)

If 5.30 g of KClO₃ decompose, how many grams of O₂ are produced?

1. Balanced Equation: 2KClO₃(s) → 2KCl(s) + 3O₂(g)
2. Molar Mass of KClO₃: K = 39.10 g/mol, Cl = 35.45 g/mol, O = 16.00 g/mol. Therefore, Molar Mass = 39.10 + (3 * 35.45) + (3 * 16.00) = 122.55 g/mol
3. Moles of KClO₃: 5.30 g KClO₃ / 122.55 g/mol = 0.0433 mol KClO₃
4. Mole Ratio: From the balanced equation, the mole ratio of KClO₃ to O₂ is 2:3. Therefore, 0.0433 mol KClO₃ * (3 mol O₂ / 2 mol KClO₃) = 0.0650 mol O₂
5. Grams of O₂: 0.0650 mol O₂ * 32.00 g/mol = 2.08 g O₂
6. Significant Figures & Units: The given mass (5.30 g) has three significant figures, so the answer should also have three significant figures. The units are grams, which is correct.

Therefore, 2.08 grams of oxygen gas are produced when 5.30 g of potassium chlorate decompose.

Example 3: Decomposition of Hydrogen Peroxide

Let's look at the decomposition of hydrogen peroxide (H₂O₂) to produce water (H₂O) and oxygen gas (O₂):

2H₂O₂(aq) → 2H₂O(l) + O₂(g)

If 10.0 g of H₂O₂ decompose, how many grams of O₂ are produced?

1. Balanced Equation: 2H₂O₂(aq) → 2H₂O(l) + O₂(g)
2. Molar Mass of H₂O₂: H = 1.01 g/mol, O = 16.00 g/mol. Therefore, M

Step 2 (continued): Molar Mass of H₂O₂
Therefore, Molar Mass = (2 × 1.01 g/mol) + (2 × 16.00 g/mol) = 2.02 g/mol + 32.00 g/mol = 34.02 g/mol.

3. Moles of H₂O₂: 10.0 g H₂O₂ / 34.02 g/mol ≈ 0.294 mol H₂O₂
4. Mole Ratio: From the balanced equation, 2 moles of H₂O₂ produce 1 mole of O₂. Thus, 0.294 mol H₂O₂ × (1 mol O₂ / 2 mol H₂O₂) = 0.147 mol O₂
5. Grams of O₂: 0.147 mol O₂ × 32.00 g/mol = 4.70 g O₂
6. Significant Figures & Units: The given mass (10.0 g) has three significant figures, so the answer is rounded to three significant figures. The units are grams, as required.

Therefore, 4.70 grams of oxygen gas are produced when 10.0 g of hydrogen peroxide decomposes.

Conclusion

The decomposition of compounds into oxygen gas, as demonstrated in these examples, relies on a systematic approach rooted in stoichiometry. By balancing chemical equations, calculating molar masses, and applying mole ratios, we can precisely determine the mass of products formed from a given reactant. This method is universally applicable to a wide range of chemical reactions, from industrial processes to laboratory experiments. The key takeaway is the importance of attention to detail—particularly in balancing equations, maintaining unit consistency, and respecting significant figures—to ensure accurate and reliable results. Mastery of these principles not only aids in solving theoretical problems but also enhances our ability to interpret and predict real-world chemical phenomena.