Verified Answer :
- Balanced equation: \(2\text{NaOH}(\text{aq}) + \text{Ni}(\text{NO}_3)_2(\text{aq}) \rightarrow \text{Ni}(\text{OH})_2(\text{s}) + 2\text{NaNO}_3(\text{aq})\).
- Calculate moles: \(\text{NaOH} = 0.8625\text{ mol}\), \(\text{Ni}(\text{NO}_3)_2 = 0.550\text{ mol}\).
- Limiting reactant is \(\text{NaOH}\) since it requires \(1.10\text{ mol}\) and only \(0.8625\text{ mol}\) is available.
- Product mass: \(\frac{0.8625}{2} \times 92.7\text{ g/mol} \approx \boxed{40.0\text{ g}}\).
Explanation
- Chemical Analysis and Balanced Equation
The reaction between sodium hydroxide (\(\text{NaOH}\)) and nickel(II) nitrate (\(\text{Ni}(\text{NO}_3)_2\)) produces nickel(II) hydroxide (\(\text{Ni}(\text{OH})_2\)) as a precipitate and sodium nitrate (\(\text{NaNO}_3\)) as an aqueous byproduct. The balanced chemical equation for this reaction is:
\[2\text{NaOH}(\text{aq}) + \text{Ni}(\text{NO}_3)_2(\text{aq}) \rightarrow \text{Ni}(\text{OH})_2(\text{s}) + 2\text{NaNO}_3(\text{aq})\]
To determine the amount of product formed, we first identify the molar masses of the substances involved, using the atomic masses: \(\text{Na}=23.0\), \(\text{O}=16.0\), \(\text{H}=1.0\), \(\text{Ni}=58.7\), and \(\text{N}=14.0\).
The molar mass of \(\text{NaOH}\) is \(23.0 + 16.0 + 1.0 = 40.0\text{ g/mol}\).
The molar mass of \(\text{Ni}(\text{NO}_3)_2\) is \(58.7 + 2 \times (14.0 + 3 \times 16.0) = 182.7\text{ g/mol}\).
The molar mass of \(\text{Ni}(\text{OH})_2\) is \(58.7 + 2 \times (16.0 + 1.0) = 92.7\text{ g/mol}\).
- Determining the Limiting Reactant
Next, we calculate the number of moles of each reactant initially present.
For \(\text{NaOH}\):
\[\text{moles of NaOH} = \frac{34.5\text{ g}}{40.0\text{ g/mol}} = 0.8625\text{ mol}\]
For \(\text{Ni}(\text{NO}_3)_2\):
\[\text{moles of Ni}(\text{NO}_3)_2 = 1.00\text{ M} \times 0.550\text{ L} = 0.550\text{ mol}\]
According to the balanced equation, 2 moles of \(\text{NaOH}\) react with 1 mole of \(\text{Ni}(\text{NO}_3)_2\). To react all \(0.550\text{ mol}\) of \(\text{Ni}(\text{NO}_3)_2\), we would need \(2 \times 0.550 = 1.10\text{ mol}\) of \(\text{NaOH}\). Since we only have \(0.8625\text{ mol}\) of \(\text{NaOH}\), \(\text{NaOH}\) is the limiting reactant.
- Calculating the Final Mass of Product
Because \(\text{NaOH}\) is the limiting reactant, the amount of product \(\text{Ni}(\text{OH})_2\) formed is determined by the amount of \(\text{NaOH}\) available. From the stoichiometry (2 moles of \(\text{NaOH}\) produce 1 mole of \(\text{Ni}(\text{OH})_2\)):
\[\text{moles of Ni}(\text{OH})_2 = \frac{0.8625\text{ mol NaOH}}{2} = 0.43125\text{ mol Ni}(\text{OH})_2\]
Finally, we convert these moles to mass using the molar mass of \(\text{Ni}(\text{OH})_2\):
\[\text{mass of Ni}(\text{OH})_2 = 0.43125\text{ mol} \times 92.7\text{ g/mol} = 39.976875\text{ g}\]
Rounding to an appropriate number of significant figures (based on the input values \(34.5\) and \(0.550\), which have 3 significant figures), the mass formed is approximately \(\boxed{40.0\text{ g}}\).
Examples
Understanding limiting reactants is crucial in industrial chemistry for minimizing waste and ensuring economic efficiency. In manufacturing, it helps engineers determine the precise amounts of raw materials required to produce a specific yield, avoiding unnecessary costs associated with surplus materials. This principle is applied in processes ranging from synthesizing pharmaceutical drugs to large-scale chemical production of materials like polymers or fertilizers.