In the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of A2B without anything left over: Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2A + B-—А,В 2A1(s) + 3Cl2 (g)→2AIC\3 (s) But what if you're given 2.8 mol of A and 3.2 mol of B? The amount of product formed is limited by the reactant that runs out first, called the limiting reactant. To identify the limiting reactant, calculate the amount of product formed from each amount of reactant separately: You are given 28.0 g of aluminum and 33.0 g of chlorine gas. Part A 2.8 metA 1 mol A2B 2 metA = 1.4 mol A2B If you had excess chlorine, how many moles of of aluminum chloride could be produced from 28.0 g of aluminum? Express your answer to three significant figures and include the appropriate units. 3.2 motB x 1 mol A,B 1 metB = 3.2 mol A2B • View Available Hint(s) Notice that less product is formed with the given amount of reactant A. Thus, A is the limiting reactant, and a maximum of 1.4 mol of A2B can be formed from the given amounts. HẢ ? Value Units
In the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of A2B without anything left over: Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2A + B-—А,В 2A1(s) + 3Cl2 (g)→2AIC\3 (s) But what if you're given 2.8 mol of A and 3.2 mol of B? The amount of product formed is limited by the reactant that runs out first, called the limiting reactant. To identify the limiting reactant, calculate the amount of product formed from each amount of reactant separately: You are given 28.0 g of aluminum and 33.0 g of chlorine gas. Part A 2.8 metA 1 mol A2B 2 metA = 1.4 mol A2B If you had excess chlorine, how many moles of of aluminum chloride could be produced from 28.0 g of aluminum? Express your answer to three significant figures and include the appropriate units. 3.2 motB x 1 mol A,B 1 metB = 3.2 mol A2B • View Available Hint(s) Notice that less product is formed with the given amount of reactant A. Thus, A is the limiting reactant, and a maximum of 1.4 mol of A2B can be formed from the given amounts. HẢ ? Value Units
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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Please answer question 10 part A and B
![### Understanding Limiting Reactants in Chemical Reactions
In a chemical reaction, reactants combine in specific proportions to form products. The limiting reactant is the one that is completely consumed first, limiting the amount of product formed.
**Example Reaction:**
\[ \text{2A + B} \rightarrow \text{A}_2\text{B} \]
#### Calculating Limiting Reactant
Suppose you have:
- 2.8 mol of A
- 3.2 mol of B
The balanced equation suggests 2 mol of A react with 1 mol of B to produce 1 mol of \(\text{A}_2\text{B}\).
1. **Calculate Product from A:**
\[
2.8 \, \cancel{\text{mol A}} \times \frac{1 \, \text{mol A}_2\text{B}}{2 \, \cancel{\text{mol A}}} = 1.4 \, \text{mol A}_2\text{B}
\]
2. **Calculate Product from B:**
\[
3.2 \, \cancel{\text{mol B}} \times \frac{1 \, \text{mol A}_2\text{B}}{1 \, \cancel{\text{mol B}}} = 3.2 \, \text{mol A}_2\text{B}
\]
Since 1.4 mol of \(\text{A}_2\text{B}\) is produced from A and 3.2 mol from B, A is the limiting reactant. Therefore, a maximum of 1.4 mol of \(\text{A}_2\text{B}\) can be formed.
### Reaction Example with Aluminum and Chlorine:
**Reaction:**
\[ \text{2Al(s) + 3Cl}_2\text{(g)} \rightarrow \text{2AlCl}_3\text{(s)} \]
**Given:**
- 28.0 g of aluminum
- 33.0 g of chlorine gas
#### Part A: Calculating Moles of \(\text{AlCl}_3\)
If chlorine is in excess, calculate the moles of \(\text{AlCl}_3\) from 28.0 g of aluminum. Express your answer to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fded90a0e-1aa2-42d3-8902-387fb21746ad%2F17b43eed-f98b-4ca9-84b3-1b4583604d3f%2Fp0foa9s_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Limiting Reactants in Chemical Reactions
In a chemical reaction, reactants combine in specific proportions to form products. The limiting reactant is the one that is completely consumed first, limiting the amount of product formed.
**Example Reaction:**
\[ \text{2A + B} \rightarrow \text{A}_2\text{B} \]
#### Calculating Limiting Reactant
Suppose you have:
- 2.8 mol of A
- 3.2 mol of B
The balanced equation suggests 2 mol of A react with 1 mol of B to produce 1 mol of \(\text{A}_2\text{B}\).
1. **Calculate Product from A:**
\[
2.8 \, \cancel{\text{mol A}} \times \frac{1 \, \text{mol A}_2\text{B}}{2 \, \cancel{\text{mol A}}} = 1.4 \, \text{mol A}_2\text{B}
\]
2. **Calculate Product from B:**
\[
3.2 \, \cancel{\text{mol B}} \times \frac{1 \, \text{mol A}_2\text{B}}{1 \, \cancel{\text{mol B}}} = 3.2 \, \text{mol A}_2\text{B}
\]
Since 1.4 mol of \(\text{A}_2\text{B}\) is produced from A and 3.2 mol from B, A is the limiting reactant. Therefore, a maximum of 1.4 mol of \(\text{A}_2\text{B}\) can be formed.
### Reaction Example with Aluminum and Chlorine:
**Reaction:**
\[ \text{2Al(s) + 3Cl}_2\text{(g)} \rightarrow \text{2AlCl}_3\text{(s)} \]
**Given:**
- 28.0 g of aluminum
- 33.0 g of chlorine gas
#### Part A: Calculating Moles of \(\text{AlCl}_3\)
If chlorine is in excess, calculate the moles of \(\text{AlCl}_3\) from 28.0 g of aluminum. Express your answer to

Transcribed Image Text:**Part B**
If you had excess aluminum, how many moles of aluminum chloride could be produced from 33.0 g of chlorine gas, Cl₂?
Express your answer to three significant figures and include the appropriate units.
- **View Available Hint(s)**
Below the text is an answer box for entering the result, labeled with "Value" and "Units." There are also several buttons related to formatting and input options, such as a periodic table symbol, reset, and keyboard icon.
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