What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M NasPO4 reacts with 42.5 mL of 0.200 M Cr(NOs)s in the following chemical reaction? NasPO«(aq) + Cr(NOs):(aq) → CIPO«(s) + 3 NaNO:(aq) |
What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M NasPO4 reacts with 42.5 mL of 0.200 M Cr(NOs)s in the following chemical reaction? NasPO«(aq) + Cr(NOs):(aq) → CIPO«(s) + 3 NaNO:(aq) |
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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![### Chemical Reaction and Precipitation Calculation
**Problem Statement**
What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Na₃PO₄ reacts with 42.5 mL of 0.200 M Cr(NO₃)₃ in the following chemical reaction?
\[ \text{Na}_3\text{PO}_4(aq) + \text{Cr(NO}_3\text{)}_3(aq) \rightarrow \text{CrPO}_4(s) + 3 \text{NaNO}_3(aq) \]
**Explanation**
In this balanced chemical equation, sodium phosphate \((\text{Na}_3\text{PO}_4)\) reacts with chromium(III) nitrate \((\text{Cr(NO}_3\text{)}_3)\) to form chromium(III) phosphate \((\text{CrPO}_4)\) as a solid precipitate and sodium nitrate \((\text{NaNO}_3)\) in aqueous solution.
To find the mass of the \(\text{CrPO}_4\) precipitate, follow these steps:
1. **Calculate moles of reactants:**
- Moles of \(\text{Na}_3\text{PO}_4 = \text{Volume (L)} \times \text{Molarity (M)}\)
- Moles of \(\text{Cr(NO}_3\text{)}_3 = \text{Volume (L)} \times \text{Molarity (M)}\)
2. **Determine the limiting reactant:**
- Use stoichiometry from the balanced equation to find which reactant forms less \(\text{CrPO}_4\).
3. **Calculate theoretical yield of \(\text{CrPO}_4\):**
- Use the moles of the limiting reactant to determine the moles of \(\text{CrPO}_4\).
- Convert moles of \(\text{CrPO}_4\) to grams using its molar mass.
This approach helps understand how reactants yield a solid product in a precipitation reaction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd862c25f-a636-4347-831e-84b629db5903%2F4b60b511-547c-4b19-988a-5b80cf9b89f4%2F5eg77_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chemical Reaction and Precipitation Calculation
**Problem Statement**
What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Na₃PO₄ reacts with 42.5 mL of 0.200 M Cr(NO₃)₃ in the following chemical reaction?
\[ \text{Na}_3\text{PO}_4(aq) + \text{Cr(NO}_3\text{)}_3(aq) \rightarrow \text{CrPO}_4(s) + 3 \text{NaNO}_3(aq) \]
**Explanation**
In this balanced chemical equation, sodium phosphate \((\text{Na}_3\text{PO}_4)\) reacts with chromium(III) nitrate \((\text{Cr(NO}_3\text{)}_3)\) to form chromium(III) phosphate \((\text{CrPO}_4)\) as a solid precipitate and sodium nitrate \((\text{NaNO}_3)\) in aqueous solution.
To find the mass of the \(\text{CrPO}_4\) precipitate, follow these steps:
1. **Calculate moles of reactants:**
- Moles of \(\text{Na}_3\text{PO}_4 = \text{Volume (L)} \times \text{Molarity (M)}\)
- Moles of \(\text{Cr(NO}_3\text{)}_3 = \text{Volume (L)} \times \text{Molarity (M)}\)
2. **Determine the limiting reactant:**
- Use stoichiometry from the balanced equation to find which reactant forms less \(\text{CrPO}_4\).
3. **Calculate theoretical yield of \(\text{CrPO}_4\):**
- Use the moles of the limiting reactant to determine the moles of \(\text{CrPO}_4\).
- Convert moles of \(\text{CrPO}_4\) to grams using its molar mass.
This approach helps understand how reactants yield a solid product in a precipitation reaction.
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