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)

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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
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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