If 29.5 g of NaOH is added to 0.700 L of 1.00 M Co(NO3)2, how many grams of Co(OH)2 will be formed in the following precipitation reaction? 2 NaOH(aq) + Co(NO3)2(aq) → Co(OH)2 (s) + 2 NaNO3(aq)
If 29.5 g of NaOH is added to 0.700 L of 1.00 M Co(NO3)2, how many grams of Co(OH)2 will be formed in the following precipitation reaction? 2 NaOH(aq) + Co(NO3)2(aq) → Co(OH)2 (s) + 2 NaNO3(aq)
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![**Chemical Reaction Calculation**
**Question 12 of 19**
If 29.5 g of NaOH is added to 0.700 L of 1.00 M Co(NO₃)₂, how many grams of Co(OH)₂ will be formed in the following precipitation reaction?
\[ 2 \text{NaOH} (aq) + \text{Co(NO}_3\text{)}_2 (aq) \rightarrow \text{Co(OH)}_2 (s) + 2 \text{NaNO}_3 (aq) \]
**Details to Consider:**
1. **Reactants and Products**:
- Sodium hydroxide (NaOH) is an aqueous solution.
- Cobalt(II) nitrate [Co(NO₃)₂] is also in an aqueous solution.
- Cobalt(II) hydroxide [Co(OH)₂] precipitates out as a solid.
- Sodium nitrate (NaNO₃) remains in an aqueous solution.
2. **Stoichiometry of the Reaction**:
- The balanced chemical equation shows that 2 moles of NaOH react with 1 mole of Co(NO₃)₂.
3. **Given Data**:
- Mass of NaOH: 29.5 g
- Volume of Co(NO₃)₂ solution: 0.700 L
- Molarity of Co(NO₃)₂ solution: 1.00 M
4. **Required Data**:
- Mass of Co(OH)₂ formed in the precipitation reaction.
**Steps to Solve**:
1. **Calculate the Number of Moles of NaOH**:
- Molar mass of NaOH = 22.99 (Na) + 15.999 (O) + 1.008 (H) = 39.997 g/mol
- Number of moles = \(\frac{29.5 g}{39.997 g/mol}\)
2. **Calculate the Number of Moles of Co(NO₃)₂ in the Solution**:
- Molarity = Moles/Volume (in L)
- Moles of Co(NO₃)₂ = 1.00 M * 0.700 L
3. **Determine the Limiting Reactant**:
- Using the stoichiometry of the reaction, compare the ratio of the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F57980898-f1ea-4baf-a646-fad54b56b73d%2F78fa590a-d7a1-46e4-89d9-a7e9ef34ddb7%2Fg7l1hh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Reaction Calculation**
**Question 12 of 19**
If 29.5 g of NaOH is added to 0.700 L of 1.00 M Co(NO₃)₂, how many grams of Co(OH)₂ will be formed in the following precipitation reaction?
\[ 2 \text{NaOH} (aq) + \text{Co(NO}_3\text{)}_2 (aq) \rightarrow \text{Co(OH)}_2 (s) + 2 \text{NaNO}_3 (aq) \]
**Details to Consider:**
1. **Reactants and Products**:
- Sodium hydroxide (NaOH) is an aqueous solution.
- Cobalt(II) nitrate [Co(NO₃)₂] is also in an aqueous solution.
- Cobalt(II) hydroxide [Co(OH)₂] precipitates out as a solid.
- Sodium nitrate (NaNO₃) remains in an aqueous solution.
2. **Stoichiometry of the Reaction**:
- The balanced chemical equation shows that 2 moles of NaOH react with 1 mole of Co(NO₃)₂.
3. **Given Data**:
- Mass of NaOH: 29.5 g
- Volume of Co(NO₃)₂ solution: 0.700 L
- Molarity of Co(NO₃)₂ solution: 1.00 M
4. **Required Data**:
- Mass of Co(OH)₂ formed in the precipitation reaction.
**Steps to Solve**:
1. **Calculate the Number of Moles of NaOH**:
- Molar mass of NaOH = 22.99 (Na) + 15.999 (O) + 1.008 (H) = 39.997 g/mol
- Number of moles = \(\frac{29.5 g}{39.997 g/mol}\)
2. **Calculate the Number of Moles of Co(NO₃)₂ in the Solution**:
- Molarity = Moles/Volume (in L)
- Moles of Co(NO₃)₂ = 1.00 M * 0.700 L
3. **Determine the Limiting Reactant**:
- Using the stoichiometry of the reaction, compare the ratio of the
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