**How to Calculate the Production of Oxygen Gas in a Chemical Reaction**In this section, we will explore a stoichiometry problem involving a chemical reaction. The goal is to determine how many grams of oxygen gas (O₂) are produced when 2.43 × 10⁻⁴ grams of potassium chlorate (KClO₃) are fully reacted.### Chemical EquationThe balanced chemical equation for the decomposition of potassium chlorate is:\[ 2 \text{KClO}_3 (s) \rightarrow 2 \text{KCl} (s) + 3 \text{O}_2 (g) \]### Problem Overview- **Given:** 2.43 × 10⁻⁴ grams of KClO₃- **Find:** Grams of O₂ produced### Calculation SetupThe interface provides boxes for entering the starting amount and various stoichiometric factors. Below are detailed steps to follow:1. Enter the starting amount of KClO₃ (2.43 × 10⁻⁴ g) in the "STARTING AMOUNT" box.2. Use the stoichiometric coefficients from the balanced equation to set up conversion factors that link moles of KClO₃ to moles of O₂.3. The molar masses needed for the conversion are: - KClO₃ = 122.55 g/mol - O₂ = 32.00 g/mol### Input Controls- **ADD FACTOR:** Add conversion factors such as the molar mass or mole ratio.- **ANSWER:** Calculate or reset to input new values.- **RESET:** Clear all input fields.### Key Concepts- **Stoichiometry:** Using mole ratios for converting between reactants and products.- **Molar Mass:** Molecular weight necessary for translating between grams and moles.By following these steps, you can accurately calculate how much oxygen is produced from a given amount of KClO₃ in this chemical reaction.

Stoichiometry involves the calculation of concentration of solutions in the given conditions of…

Answered: How many grams of oxygen gas are…

How many grams of oxygen gas are produced when 2.43 x 10-g of KCIO: are completely reacted according to the following chemical equation: 2 KCIO:(s) – 2 KC(s) + 3 o:(g)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

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Concept explainers

Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

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**How to Calculate the Production of Oxygen Gas in a Chemical Reaction**

In this section, we will explore a stoichiometry problem involving a chemical reaction. The goal is to determine how many grams of oxygen gas (O₂) are produced when 2.43 × 10⁻⁴ grams of potassium chlorate (KClO₃) are fully reacted.

### Chemical Equation

The balanced chemical equation for the decomposition of potassium chlorate is:

\[ 2 \text{KClO}_3 (s) \rightarrow 2 \text{KCl} (s) + 3 \text{O}_2 (g) \]

### Problem Overview

- **Given:** 2.43 × 10⁻⁴ grams of KClO₃
- **Find:** Grams of O₂ produced

### Calculation Setup

The interface provides boxes for entering the starting amount and various stoichiometric factors. Below are detailed steps to follow:

1. Enter the starting amount of KClO₃ (2.43 × 10⁻⁴ g) in the "STARTING AMOUNT" box.
2. Use the stoichiometric coefficients from the balanced equation to set up conversion factors that link moles of KClO₃ to moles of O₂.
3. The molar masses needed for the conversion are:
- KClO₃ = 122.55 g/mol
- O₂ = 32.00 g/mol

### Input Controls

- **ADD FACTOR:** Add conversion factors such as the molar mass or mole ratio.
- **ANSWER:** Calculate or reset to input new values.
- **RESET:** Clear all input fields.

### Key Concepts

- **Stoichiometry:** Using mole ratios for converting between reactants and products.
- **Molar Mass:** Molecular weight necessary for translating between grams and moles.

By following these steps, you can accurately calculate how much oxygen is produced from a given amount of KClO₃ in this chemical reaction.

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