What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Na3PO4 reacts with 58.0 mL of 0.200 M Cr(NO3)3 in the following chemical reaction? Na3PO4 (aq) + Cr(NO3)3(aq) CrPO4(s) + 3 NaNO3(aq)
What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Na3PO4 reacts with 58.0 mL of 0.200 M Cr(NO3)3 in the following chemical reaction? Na3PO4 (aq) + Cr(NO3)3(aq) CrPO4(s) + 3 NaNO3(aq)
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Stoichiometry Problem: Calculating Mass of Precipitate**
**Question:**
What mass of precipitate (in grams) is formed when 45.5 mL of 0.300 M Na₃PO₄ reacts with 58.0 mL of 0.200 M Cr(NO₃)₃ in the following chemical reaction?
\[
\text{Na}_3\text{PO}_4 (\text{aq}) + \text{Cr}(\text{NO}_3)_3 (\text{aq}) \rightarrow \text{CrPO}_4 (\text{s}) + 3 \text{NaNO}_3 (\text{aq})
\]
**Solution:**
1. **Determine the moles of each reactant:**
- Na₃PO₄: \( 45.5 \, \text{mL} \times 0.300 \, \text{mol/L} = 0.01365 \, \text{mol} \)
- Cr(NO₃)₃: \( 58.0 \, \text{mL} \times 0.200 \, \text{mol/L} = 0.0116 \, \text{mol} \)
2. **Identify the limiting reactant:**
- From the balanced equation, the mole ratio of Na₃PO₄ to CrPO₄ is 1:1.
- Cr(NO₃)₃ has fewer moles, so it is the limiting reactant.
3. **Calculate the mass of CrPO₄ (precipitate):**
- Moles of CrPO₄ produced = 0.0116 mol (from Cr(NO₃)₃)
- Molar mass of CrPO₄ = 146.00 g/mol
- Mass = moles × molar mass = 0.0116 mol × 146.00 g/mol = 1.6936 g
Therefore, 1.6936 grams of CrPO₄ precipitate are formed.
**Note:** Remember to always verify the stoichiometry and the limiting reactant when performing calculations involving chemical reactions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2ebf2b47-0653-48b1-aa4f-9532848d4742%2F054cec03-3936-406d-9795-0da9e255f710%2Fstxzyyj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Stoichiometry Problem: Calculating Mass of Precipitate**
**Question:**
What mass of precipitate (in grams) is formed when 45.5 mL of 0.300 M Na₃PO₄ reacts with 58.0 mL of 0.200 M Cr(NO₃)₃ in the following chemical reaction?
\[
\text{Na}_3\text{PO}_4 (\text{aq}) + \text{Cr}(\text{NO}_3)_3 (\text{aq}) \rightarrow \text{CrPO}_4 (\text{s}) + 3 \text{NaNO}_3 (\text{aq})
\]
**Solution:**
1. **Determine the moles of each reactant:**
- Na₃PO₄: \( 45.5 \, \text{mL} \times 0.300 \, \text{mol/L} = 0.01365 \, \text{mol} \)
- Cr(NO₃)₃: \( 58.0 \, \text{mL} \times 0.200 \, \text{mol/L} = 0.0116 \, \text{mol} \)
2. **Identify the limiting reactant:**
- From the balanced equation, the mole ratio of Na₃PO₄ to CrPO₄ is 1:1.
- Cr(NO₃)₃ has fewer moles, so it is the limiting reactant.
3. **Calculate the mass of CrPO₄ (precipitate):**
- Moles of CrPO₄ produced = 0.0116 mol (from Cr(NO₃)₃)
- Molar mass of CrPO₄ = 146.00 g/mol
- Mass = moles × molar mass = 0.0116 mol × 146.00 g/mol = 1.6936 g
Therefore, 1.6936 grams of CrPO₄ precipitate are formed.
**Note:** Remember to always verify the stoichiometry and the limiting reactant when performing calculations involving chemical reactions.
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