Question 9 of 18 What mass of precipitate (ing) is formed when 45.5 mL of 0.300 M K₂PO4 reacts with 36.0 mL of 0.200 M Cr(NO3)3 in the following chemical reaction? K3PO4(aq) + Cr(NO3)3(aq) → CrPO4(s) + 3 KNO3(aq) 1 4 7 +/- Tap here or pull up for additional resources 25 8 3 6 O) 9 g с XU Submit x 100
Question 9 of 18 What mass of precipitate (ing) is formed when 45.5 mL of 0.300 M K₂PO4 reacts with 36.0 mL of 0.200 M Cr(NO3)3 in the following chemical reaction? K3PO4(aq) + Cr(NO3)3(aq) → CrPO4(s) + 3 KNO3(aq) 1 4 7 +/- Tap here or pull up for additional resources 25 8 3 6 O) 9 g с XU Submit x 100
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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![**Example Problem: Calculating Mass of Precipitate in a Chemical Reaction**
**Question:**
What mass of precipitate (in grams) is formed when 45.5 mL of 0.300 M K₃PO₄ reacts with 36.0 mL of 0.200 M Cr(NO₃)₃ in the following chemical reaction?
\[ \text{K}_3\text{PO}_4(aq) + \text{Cr(NO}_3\text{)}_3(aq) \rightarrow \text{CrPO}_4(s) + 3 \text{KNO}_3(aq) \]
**Solution Steps:**
1. **Determine the moles of reactants:**
- Calculate moles of \( \text{K}_3\text{PO}_4 \):
\[
\text{Moles of } \text{K}_3\text{PO}_4 = 0.300 \, \text{M} \times 0.0455 \, \text{L} = 0.01365 \, \text{moles}
\]
- Calculate moles of \( \text{Cr(NO}_3\text{)}_3 \):
\[
\text{Moles of } \text{Cr(NO}_3\text{)}_3 = 0.200 \, \text{M} \times 0.0360 \, \text{L} = 0.0072 \, \text{moles}
\]
2. **Determine the limiting reactant:**
- According to the balanced equation, 1 mole of \( \text{K}_3\text{PO}_4 \) reacts with 1 mole of \( \text{Cr(NO}_3\text{)}_3 \).
- \( \text{Cr(NO}_3\text{)}_3 \) is the limiting reactant because it has fewer moles (0.0072 moles).
3. **Calculate moles of precipitate, \( \text{CrPO}_4 \):**
- Since \( \text{Cr(NO}_3\text{)}_3 \) is the limiting reactant, the moles of \( \text{CrPO}_4 \) formed will be equal to the moles of \( \text{Cr](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F38543b38-0ca2-4898-a226-639f5edfcbc9%2Ffa31eb6d-1038-4404-9cff-6307c19d035c%2Fvxdo25n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Example Problem: Calculating Mass of Precipitate in a Chemical Reaction**
**Question:**
What mass of precipitate (in grams) is formed when 45.5 mL of 0.300 M K₃PO₄ reacts with 36.0 mL of 0.200 M Cr(NO₃)₃ in the following chemical reaction?
\[ \text{K}_3\text{PO}_4(aq) + \text{Cr(NO}_3\text{)}_3(aq) \rightarrow \text{CrPO}_4(s) + 3 \text{KNO}_3(aq) \]
**Solution Steps:**
1. **Determine the moles of reactants:**
- Calculate moles of \( \text{K}_3\text{PO}_4 \):
\[
\text{Moles of } \text{K}_3\text{PO}_4 = 0.300 \, \text{M} \times 0.0455 \, \text{L} = 0.01365 \, \text{moles}
\]
- Calculate moles of \( \text{Cr(NO}_3\text{)}_3 \):
\[
\text{Moles of } \text{Cr(NO}_3\text{)}_3 = 0.200 \, \text{M} \times 0.0360 \, \text{L} = 0.0072 \, \text{moles}
\]
2. **Determine the limiting reactant:**
- According to the balanced equation, 1 mole of \( \text{K}_3\text{PO}_4 \) reacts with 1 mole of \( \text{Cr(NO}_3\text{)}_3 \).
- \( \text{Cr(NO}_3\text{)}_3 \) is the limiting reactant because it has fewer moles (0.0072 moles).
3. **Calculate moles of precipitate, \( \text{CrPO}_4 \):**
- Since \( \text{Cr(NO}_3\text{)}_3 \) is the limiting reactant, the moles of \( \text{CrPO}_4 \) formed will be equal to the moles of \( \text{Cr
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