Question

Consider the precipitation reaction: $$2 \mathrm{Na}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{CuCl}_{2}(a q) \longrightarrow \mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{NaCl}(a q)$$ What volume of 0.175 \(\mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution is necessary to completely react with 95.4 \(\mathrm{mL}\) of 0.102 \(\mathrm{M} \mathrm{CuCl}_{2} ?\)

Short answer

To completely react with 95.4 mL of 0.102 M CuCl2, the volume of 0.175 M Na3PO4 needed is calculated using stoichiometry based on the balanced chemical reaction.

Step-by-step solution

Write a balanced chemical equation

The balanced chemical equation for this reaction is already provided: $$2 \mathrm{Na}_3\mathrm{PO}_4(aq) + 3 \mathrm{CuCl}_2(aq) \longrightarrow \mathrm{Cu}_3(\mathrm{PO}_4)_2(s) + 6 \mathrm{NaCl}(aq)$$This equation indicates that 2 moles of \(\mathrm{Na}_3\mathrm{PO}_4\) react with 3 moles of \(\mathrm{CuCl}_2\) to form products.

Calculate moles of CuCl2

Use the concentration and volume of \(\mathrm{CuCl}_2\) to find the moles of \(\mathrm{CuCl}_2\) that will react.$$\text{Moles of } \mathrm{CuCl}_2 = \text{Volume (in L) } \times \text{ Molarity (in mol/L) }$$Convert the volume of \(\mathrm{CuCl}_2\) from mL to L:$$95.4 \, \text{mL} = 0.0954 \, \text{L}$$Now calculate the moles:$$\text{Moles of }\mathrm{CuCl}_2 = 0.0954 \, \text{L} \times 0.102 \, \text{mol/L}$$

Calculate the moles of Na3PO4

Using the stoichiometry from the balanced chemical equation, calculate the moles of \(\mathrm{Na}_3\mathrm{PO}_4\) needed to react with \(\mathrm{CuCl}_2\).$$\frac{2 \, \text{moles of } \mathrm{Na}_3\mathrm{PO}_4}{3 \, \text{moles of } \mathrm{CuCl}_2} = \frac{\text{Moles of } \mathrm{Na}_3\mathrm{PO}_4}{\text{Moles of } \mathrm{CuCl}_2}$$Solve this proportion for the moles of \(\mathrm{Na}_3\mathrm{PO}_4\):

Calculate the volume of Na3PO4 needed

Now, calculate the volume of the \(0.175\, \mathrm{M}\) \(\mathrm{Na}_3\mathrm{PO}_4\) solution required by using the number of moles and its molarity.$$\text{Volume of } \mathrm{Na}_3\mathrm{PO}_4\, (\text{in L}) = \frac{\text{Moles of } \mathrm{Na}_3\mathrm{PO}_4}{\text{Molarity of } \mathrm{Na}_3\mathrm{PO}_4\, (\text{in mol/L})}$$

Key concepts

Stoichiometry

Stoichiometry is the section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. It is a key part of understanding chemical reactions and is essential for carrying out chemical calculations.

In our example, stoichiometry is used to relate the amounts of sodium phosphate \(\mathrm{Na}_3\mathrm{PO}_4\) and copper(II) chloride \(\mathrm{CuCl}_2\) in the precipitation reaction. By knowing the balanced chemical equation, we can determine the proportion of reactants that combine to form the products.

For instance, the balanced equation shows a 2:3 mole ratio, which means for every 2 moles of sodium phosphate used, 3 moles of copper(II) chloride are needed. This ratio is paramount when we want to calculate the amounts of each reactant necessary for the reaction to go to completion without leaving an excess of one over the other. Getting the stoichiometry right ensures that we make accurate and efficient use of materials in chemical processes.

Molarity Calculations

Molarity calculations are essential for determining the concentration of a solution, which is expressed as the number of moles of solute per liter of solution (mol/L). It helps chemists to prepare solutions of precise concentrations and to carry out reactions with the aforementioned controlled proportions.

In the exercise's context, the molarity of the copper(II) chloride solution is given (0.102 M), and it's used to figure out the amount of substance, in moles, that we have in a certain volume of solution. The calculation is made by multiplying the volume of the solution (expressed in liters) by its molarity. This step is crucial in determining how much of the sodium phosphate solution is required because it directly tells us how many moles of copper(II) chloride we have to start with, which in turn dictates how much sodium phosphate is needed according to the stoichiometric ratio.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is respected; matter cannot be created or destroyed in a chemical reaction. To 'balance' an equation, one adjusts the coefficients (the numbers in front of the chemical formulas) to make the number of atoms of each element the same on both sides of the equation.

The balanced equation provided in the exercise, \(2 \mathrm{Na}_3\mathrm{PO}_4(aq) + 3 \mathrm{CuCl}_2(aq) \longrightarrow \mathrm{Cu}_3(\mathrm{PO}_4)_2(s) + 6 \mathrm{NaCl}(aq)\), indicates the exact proportions in which reactants combine and the quantities in which products form. Balancing equations is the initial step that guides us through the rest of the calculation process. In teaching chemical equation balancing, it's important to suggest working methodically, perhaps by balancing elements that appear in the fewest compounds first, and leaving hydrogen and oxygen for last.