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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![**Problem: Determining Moles of Iron(III) Nitrate**
How many moles of iron(III) nitrate, Fe(NO₃)₃, are there in 199 mL of a 0.136 M solution?
[Input Box] moles Fe(NO₃)₃
**Explanation:**
In this problem, you are required to determine the amount of moles of iron(III) nitrate, Fe(NO₃)₃, present in a given volume of solution. The volume of the solution is given in milliliters (199 mL), and the concentration of the solution is given in molarity (0.136 M).
**Steps to Calculate Moles:**
1. **Convert the Volume to Liters:** Since molarity (M) is defined as moles per liter, we need to convert the volume from milliliters to liters.
\[ 199 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.199 \text{ L} \]
2. **Use the Molarity Formula:**
\[ \text{Molarity (M)} = \frac{\text{Moles of Solute}}{\text{Volume of Solution in Liters}} \]
We can rearrange this formula to solve for the moles of solute:
\[ \text{Moles of Fe(NO₃)₃} = \text{Molarity (M)} \times \text{Volume (L)} \]
3. **Calculate the Moles of Fe(NO₃)₃:**
\[ \text{Moles of Fe(NO₃)₃} = 0.136 \text{ M} \times 0.199 \text{ L} = 0.027064 \text{ moles} \]
Therefore, you will enter the value 0.027064 in the input box for moles of Fe(NO₃)₃.
This calculation shows that there are approximately 0.0271 moles of iron(III) nitrate in a 199 mL solution with a molarity of 0.136 M.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa527ffdc-16df-4a1f-9953-96956a75d05e%2F3d788191-5aa1-4b10-ac92-4c94abb1b52e%2Feudk9jg.png&w=3840&q=75)
Transcribed Image Text:**Problem: Determining Moles of Iron(III) Nitrate**
How many moles of iron(III) nitrate, Fe(NO₃)₃, are there in 199 mL of a 0.136 M solution?
[Input Box] moles Fe(NO₃)₃
**Explanation:**
In this problem, you are required to determine the amount of moles of iron(III) nitrate, Fe(NO₃)₃, present in a given volume of solution. The volume of the solution is given in milliliters (199 mL), and the concentration of the solution is given in molarity (0.136 M).
**Steps to Calculate Moles:**
1. **Convert the Volume to Liters:** Since molarity (M) is defined as moles per liter, we need to convert the volume from milliliters to liters.
\[ 199 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.199 \text{ L} \]
2. **Use the Molarity Formula:**
\[ \text{Molarity (M)} = \frac{\text{Moles of Solute}}{\text{Volume of Solution in Liters}} \]
We can rearrange this formula to solve for the moles of solute:
\[ \text{Moles of Fe(NO₃)₃} = \text{Molarity (M)} \times \text{Volume (L)} \]
3. **Calculate the Moles of Fe(NO₃)₃:**
\[ \text{Moles of Fe(NO₃)₃} = 0.136 \text{ M} \times 0.199 \text{ L} = 0.027064 \text{ moles} \]
Therefore, you will enter the value 0.027064 in the input box for moles of Fe(NO₃)₃.
This calculation shows that there are approximately 0.0271 moles of iron(III) nitrate in a 199 mL solution with a molarity of 0.136 M.
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