**Title: Exploring the Use of Acetylene Gas in High Temperature Torches**Acetylene gas, \( \text{HCCH} \), is commonly used in high temperature torches. Understanding the chemical reactions it undergoes is crucial for both safety and efficiency. This section explores one such reaction with hydrogen gas to form ethane.**a. Balanced Chemical Equation**To form ethane (\( \text{C}_2\text{H}_6 \)), acetylene reacts with hydrogen gas. The balanced chemical equation for this reaction is:\[ \text{C}_2\text{H}_2 + 2\text{H}_2 \rightarrow \text{C}_2\text{H}_6 \]This equation shows that one molecule of acetylene reacts with two molecules of hydrogen to produce one molecule of ethane.**b. Calculating Theoretical Yield and Limiting Reactant**To determine how many grams of ethane can be produced from a mixture of 30.3 g of acetylene and 4.14 g of hydrogen, we need to consider the limiting reactant and theoretical yield of this reaction.1. **Molar Mass Calculation**: - Molar mass of acetylene (\(\text{C}_2\text{H}_2\)): \(26.04 \, \text{g/mol}\) - Molar mass of hydrogen (\(\text{H}_2\)): \(2.02 \, \text{g/mol}\) - Molar mass of ethane (\(\text{C}_2\text{H}_6\)): \(30.07 \, \text{g/mol}\)2. **Moles Calculation**: - Moles of acetylene: \(\frac{30.3 \, \text{g}}{26.04 \, \text{g/mol}} = 1.164 \, \text{mol}\) - Moles of hydrogen: \(\frac{4.14 \, \text{g}}{2.02 \, \text{g/mol}} = 2.050 \, \text{mol}\)3. **Limiting Reactant Determination**: - According to the balanced equation, 2 moles of hydrogen are required for each mole of acetylene. - Required moles of hydrogen for 1.164 mol of acetylene: \(2 \times

a. The balanced chemical reaction of acetylene with hydrogen gas to form ethane is given as, C2H2…

Acetylene gas, HCCH, is commonly used in high temperature torches. a. Write a balanced chemical equation for the reaction of acetylene with hydrogen gas to form ethane (C;Hs). b. How many grams of ethane can be produced from a mixture of 30.3 g acetylene and 4.14 g of hydrogen? Which reactant is limiting and what is the theoretical yield?

Acetylene gas, HCCH, is commonly used in high temperature torches. a. Write a balanced chemical equation for the reaction of acetylene with hydrogen gas to form ethane (C;Hs). b. How many grams of ethane can be produced from a mixture of 30.3 g acetylene and 4.14 g of hydrogen? Which reactant is limiting and what is the theoretical yield?

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

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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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".

Question

**Title: Exploring the Use of Acetylene Gas in High Temperature Torches**

Acetylene gas, \( \text{HCCH} \), is commonly used in high temperature torches. Understanding the chemical reactions it undergoes is crucial for both safety and efficiency. This section explores one such reaction with hydrogen gas to form ethane.

**a. Balanced Chemical Equation**

To form ethane (\( \text{C}_2\text{H}_6 \)), acetylene reacts with hydrogen gas. The balanced chemical equation for this reaction is:

\[ \text{C}_2\text{H}_2 + 2\text{H}_2 \rightarrow \text{C}_2\text{H}_6 \]

This equation shows that one molecule of acetylene reacts with two molecules of hydrogen to produce one molecule of ethane.

**b. Calculating Theoretical Yield and Limiting Reactant**

To determine how many grams of ethane can be produced from a mixture of 30.3 g of acetylene and 4.14 g of hydrogen, we need to consider the limiting reactant and theoretical yield of this reaction.

1. **Molar Mass Calculation**:
- Molar mass of acetylene (\(\text{C}_2\text{H}_2\)): \(26.04 \, \text{g/mol}\)
- Molar mass of hydrogen (\(\text{H}_2\)): \(2.02 \, \text{g/mol}\)
- Molar mass of ethane (\(\text{C}_2\text{H}_6\)): \(30.07 \, \text{g/mol}\)

2. **Moles Calculation**:
- Moles of acetylene: \(\frac{30.3 \, \text{g}}{26.04 \, \text{g/mol}} = 1.164 \, \text{mol}\)
- Moles of hydrogen: \(\frac{4.14 \, \text{g}}{2.02 \, \text{g/mol}} = 2.050 \, \text{mol}\)

3. **Limiting Reactant Determination**:
- According to the balanced equation, 2 moles of hydrogen are required for each mole of acetylene.
- Required moles of hydrogen for 1.164 mol of acetylene: \(2 \times

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