Verified Answer :
- Calculate the molar masses of \(\text{NaOH}\) (39.998 g/mol) and \(\text{Ni}(\text{OH})_2\) (92.706 g/mol).
- Determine the moles of reactants: 0.8625 mol of \(\text{NaOH}\) and 0.550 mol of \(\text{Ni}(\text{NO}_3)_2\).
- Identify \(\text{NaOH}\) as the limiting reactant since it requires 1.100 mol to react with all the nickel salt.
- Calculate the mass of product based on \(\text{NaOH}\): \(\frac{0.8625}{2} \times 92.706 \approx 39.98 \text{ g}\).
- Final Answer: \(\boxed{39.98 \text{ g}}\).
Explanation
- Analysis and Molar Masses
To determine the amount of nickel(II) hydroxide, \(\text{Ni}(\text{OH})_2\), formed, we must first analyze the stoichiometry of the precipitation reaction:
\[2\text{NaOH}(aq) + \text{Ni}(\text{NO}_3)_2(aq) \to \text{Ni}(\text{OH})_2(s) + 2\text{NaNO}_3(aq)\]
We are given 34.5 g of \(\text{NaOH}\) and 0.550 L of a 1.00 M \(\text{Ni}(\text{NO}_3)_2\) solution. To perform the stoichiometry, we first calculate the molar masses of the substances involved using atomic masses:
- For \(\text{NaOH}\): \(22.99 (\text{Na}) + 16.00 (\text{O}) + 1.008 (\text{H}) = 39.998 \text{ g/mol}\)
- For \(\text{Ni}(\text{OH})_2\): \(58.69 (\text{Ni}) + 2 \times (16.00 + 1.008) = 58.69 + 34.016 = 92.706 \text{ g/mol}\)
- Identifying the Limiting Reactant
Next, we convert the given quantities into moles:
- Moles of \(\text{NaOH}\): \(n_{\text{NaOH}} = \frac{34.5 \text{ g}}{39.998 \text{ g/mol}} \approx 0.8625 \text{ mol}\)
- Moles of \(\text{Ni}(\text{NO}_3)_2\): \(n_{\text{Ni}(\text{NO}_3)_2} = 1.00 \text{ mol/L} \times 0.550 \text{ L} = 0.550 \text{ mol}\)
To identify the limiting reactant, we look at the balanced chemical equation, which requires 2 moles of \(\text{NaOH}\) for every 1 mole of \(\text{Ni}(\text{NO}_3)_2\). If we were to use all 0.550 moles of \(\text{Ni}(\text{NO}_3)_2\), we would need \(0.550 \times 2 = 1.100 \text{ moles of NaOH}\). Since we have only 0.8625 moles of \(\text{NaOH}\), \(\text{NaOH}\) is clearly the limiting reactant.
- Calculating the Mass of Product
Since \(\text{NaOH}\) is the limiting reactant, the amount of product formed is determined by the amount of \(\text{NaOH}\) available. The balanced equation shows a 2:1 stoichiometric ratio between \(\text{NaOH}\) and \(\text{Ni}(\text{OH})_2\).
- Moles of \(\text{Ni}(\text{OH})_2\) produced: \(n_{\text{Ni}(\text{OH})_2} = \frac{0.8625 \text{ mol NaOH}}{2} = 0.43125 \text{ mol}\)
Finally, we calculate the mass of \(\text{Ni}(\text{OH})_2\) formed:
- Mass = \(n_{\text{Ni}(\text{OH})_2} \times M_{\text{Ni}(\text{OH})_2} = 0.43125 \text{ mol} \times 92.706 \text{ g/mol} \approx 39.98 \text{ g}\).
Thus, the mass of \(\text{Ni}(\text{OH})_2\) formed is approximately \(\boxed{39.98 \text{ g}}\).
Examples
Understanding limiting reactants is essential in industrial chemistry and laboratory settings where reagents are expensive or limited. For instance, in manufacturing pharmaceuticals, you want to ensure that expensive precursors are used as efficiently as possible, or that you correctly calculate the yield to minimize waste. By identifying the limiting reagent, chemists can optimize reaction conditions and predict product yields, ensuring economic and environmental efficiency in chemical processes.