According to the following reaction, how many grams of ethylene (C,H4) are required for the complete reaction of 30.3 grams of hydrogen gas? hydrogen (g) + ethylene (C,H4) (g) ethane (C,H,) grams ethylene (C,H4)
According to the following reaction, how many grams of ethylene (C,H4) are required for the complete reaction of 30.3 grams of hydrogen gas? hydrogen (g) + ethylene (C,H4) (g) ethane (C,H,) grams ethylene (C,H4)
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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**Calculating the Required Mass of Ethylene for a Chemical Reaction**
In a chemical reaction, stoichiometry allows us to calculate the amounts of reactants needed or products formed. Here, we need to determine the mass of ethylene (\(C_2H_4\)) required to completely react with a given amount of hydrogen gas.
**Reaction Equation:**
\[ \text{hydrogen} (g) + \text{ethylene} (C_2H_4) (g) \rightarrow \text{ethane} (C_2H_6) (g) \]
**Problem Statement:**
How many grams of ethylene (\(C_2H_4\)) are needed for the complete reaction of 30.3 grams of hydrogen gas?
**Procedure:**
1. **Determine the Molar Mass:**
- Hydrogen (\(H_2\)): \(2 \times 1.01 \, \text{g/mol} = 2.02 \, \text{g/mol}\)
- Ethylene (\(C_2H_4\)): \(2 \times 12.01 + 4 \times 1.01 \, \text{g/mol} = 28.05 \, \text{g/mol}\)
2. **Calculate Moles of Hydrogen:**
\[ \text{Moles of } H_2 = \frac{30.3 \, \text{g}}{2.02 \, \text{g/mol}} \]
3. **Use Stoichiometry:**
- The balanced equation indicates the molar ratio between hydrogen and ethylene.
- Based on the reaction, \(1 \text{ mole of } H_2\) reacts with \(1 \text{ mole of } C_2H_4\).
4. **Calculate Mass of Ethylene:**
\[ \text{Mass of } C_2H_4 = \text{Moles of } C_2H_4 \times 28.05 \, \text{g/mol} \]
**Calculate and Fill in the Blank:**
\[ \text{grams ethylene} (C_2H_4) \]
This process highlights the fundamental application of stoichiometry in determining the quantities of reactants required for a chemical reaction to proceed completely.
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Transcribed Image Text:---
**Calculating the Required Mass of Ethylene for a Chemical Reaction**
In a chemical reaction, stoichiometry allows us to calculate the amounts of reactants needed or products formed. Here, we need to determine the mass of ethylene (\(C_2H_4\)) required to completely react with a given amount of hydrogen gas.
**Reaction Equation:**
\[ \text{hydrogen} (g) + \text{ethylene} (C_2H_4) (g) \rightarrow \text{ethane} (C_2H_6) (g) \]
**Problem Statement:**
How many grams of ethylene (\(C_2H_4\)) are needed for the complete reaction of 30.3 grams of hydrogen gas?
**Procedure:**
1. **Determine the Molar Mass:**
- Hydrogen (\(H_2\)): \(2 \times 1.01 \, \text{g/mol} = 2.02 \, \text{g/mol}\)
- Ethylene (\(C_2H_4\)): \(2 \times 12.01 + 4 \times 1.01 \, \text{g/mol} = 28.05 \, \text{g/mol}\)
2. **Calculate Moles of Hydrogen:**
\[ \text{Moles of } H_2 = \frac{30.3 \, \text{g}}{2.02 \, \text{g/mol}} \]
3. **Use Stoichiometry:**
- The balanced equation indicates the molar ratio between hydrogen and ethylene.
- Based on the reaction, \(1 \text{ mole of } H_2\) reacts with \(1 \text{ mole of } C_2H_4\).
4. **Calculate Mass of Ethylene:**
\[ \text{Mass of } C_2H_4 = \text{Moles of } C_2H_4 \times 28.05 \, \text{g/mol} \]
**Calculate and Fill in the Blank:**
\[ \text{grams ethylene} (C_2H_4) \]
This process highlights the fundamental application of stoichiometry in determining the quantities of reactants required for a chemical reaction to proceed completely.
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