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How many moles of iron(III) nitrate, Fe(NO3)3, are there in 199 mL of a 0.136 M solution?

How many moles of iron(III) nitrate, Fe(NO3)3, are there in 199 mL of a 0.136 M solution?

Chemistry
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.
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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