Question

How many milliliters of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}\) solution are required to precipitate all the nickel, as NiS, from 25.0 \(\mathrm{mL}\) of \(0.20 \mathrm{M} \mathrm{NiCl}_{2}\) solution? \(\mathrm{NiCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{S}(a q) \rightarrow \mathrm{NiS}(s)+2 \mathrm{NaCl}(a q)\)

Short answer

To precipitate all the nickel from 25.0 mL of \(0.20 \mathrm{M} \mathrm{NiCl}_{2}\) solution, 50.0 mL of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}\) solution is required.

Step-by-step solution

Understanding the chemical equation

The given chemical equation is as follows: \[ \mathrm{NiCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{S}(a q) \rightarrow \mathrm{NiS}(s)+2 \mathrm{NaCl}(a q) \] This equation tells us that 1 mole of \(\mathrm{NiCl}_{2}\) reacts with 1 mole of \(\mathrm{Na}_{2} \mathrm{S}\) to form 1 mole of \(\mathrm{NiS}\) and 2 moles of \(\mathrm{NaCl}\). Thus, the molar ratio between \(\mathrm{NiCl}_{2}\) and \(\mathrm{Na}_{2} \mathrm{S}\) is 1:1.

Determine the moles of \(\mathrm{NiCl}_{2}\)

Next, calculate the moles of \(\mathrm{NiCl}_{2}\). This can be done using the formula for molarity which is moles of solute over liters of solution. Hence the moles of \(\mathrm{NiCl}_{2}\) can be found as follows: \[ \text{Moles of }\mathrm{NiCl}_{2} = Molarity \times Volume \] By substituting the given values, \[ \text{Moles of }\mathrm{NiCl}_{2} = 0.20 \, \mathrm{M} \times 0.025 \, \mathrm{L} \] which gives 0.005 moles of \(\mathrm{NiCl}_{2}\).

Determine the moles of \(\mathrm{Na}_{2} \mathrm{S}\) required

Using the stoichiometric ratio from step 1, we can set up a ratio to find the amount of \(\mathrm{Na}_{2} \mathrm{S}\) needed to react with \(\mathrm{NiCl}_{2}\). Since the ratio is 1:1, we need the same amount of moles of \(\mathrm{Na}_{2} \mathrm{S}\) to react with \(\mathrm{NiCl}_{2}\). Therefore, 0.005 moles of \(\mathrm{Na}_{2} \mathrm{S}\) are needed.

Calculate the Volume of \(\mathrm{Na}_{2} \mathrm{S}\) required

To determine how much volume this amount of moles occupies, use the molarity formula again. In this case, we solve for volume: \[ \text{Volume} = \frac{\text{Moles of solute}}{\text{Molarity}} \] Substituting the calculated moles and given molarity, \[ \text{Volume} = \frac{0.005\, \mathrm{mol}}{0.10\, \mathrm{M}} \] This gives a result of 0.05 Liters or 50.0 mL. So, 50.0 mL of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}\) solution is required to precipitate all the nickel from 25.0 mL of \(0.20 \mathrm{M} \mathrm{NiCl}_{2}\) solution.

Key concepts

Molarity

Understanding molarity allows us to determine the concentration of a solution in terms of moles per liter (mol/L). It is a useful tool in stoichiometry for converting volume measurements into moles, which are needed for chemical calculations.

• Molarity is expressed as \( M \), representing moles of solute per liter of solution.
• The formula is given by: \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \).
• In our exercise, the molarity of \( \mathrm{NiCl}_2 \) is utilized to determine the number of moles present in a 25.0 mL solution, thus helping in the calculation of the required amount of \( \mathrm{Na}_2 \mathrm{S} \) to complete the reaction.

By grasping the concept of molarity, you can easily switch between measuring volume in liters and counting moles, thereby simplifying the stoichiometric computations in reactions.

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely consumed first, thereby stopping the reaction from continuing further.

• Determining the limiting reactant is crucial because it dictates the amount of product that can be formed during a chemical reaction.
• When two reactants are combined, a stoichiometric calculation is performed to find out which reactant will run out first. This is accomplished by converting all reactants into moles and comparing their stoichiometric ratios defined by the chemical equation.
• In our example, the stoichiometric ratio of \( \mathrm{NiCl}_2 \) to \( \mathrm{Na}_2 \mathrm{S} \) is 1:1, meaning one mole of each reactant is necessary to produce one mole of \( \mathrm{NiS} \).

By calculating the moles available and required, the method allows students to understand the concept of a limiting reactant and its importance in determining the extent and outcomes of chemical reactions.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through bond breaking and forming.

• Each chemical reaction has a balanced equation that indicates which and how many reactants combine to form specific amounts of products.
• Understanding chemical equations allows chemists to predict the outcome of reactions and the quantities of materials needed or produced.
• In our scenario, the equation \( \mathrm{NiCl}_2(aq) + \mathrm{Na}_2 \mathrm{S}(aq) \rightarrow \mathrm{NiS}(s) + 2 \mathrm{NaCl}(aq) \) shows that the reaction will produce nickel sulfide, \( \mathrm{NiS} \), a solid precipitate from the solutions provided.

Studying chemical reactions and their equations equips learners with the ability to follow such processes meticulously, predict the identity and quantity of products, and understand the chemical and physical changes inherent in reactions.