Question

A student prepares phosphorous acid, \(\mathrm{H}_{3} \mathrm{PO}_{3}\), by reacting solid phosphorus triiodide with water. $$ \mathrm{PI}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(s)+3 \mathrm{HI}(g) $$ The student needs to obtain \(0.250 \mathrm{~L}\) of \(\mathrm{H}_{3} \mathrm{PO}_{3}(d=1.651\) \(\left.\mathrm{g} / \mathrm{cm}^{3}\right)\). The procedure calls for a \(45.0 \%\) excess of water and a yield of \(75.0 \% .\) How much phosphorus triiodide should be weighed out? What volume of water \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\) should be used?

Short answer

Answer: To obtain 0.250 liters of phosphorous acid, 2763.53 grams of phosphorus triiodide and 0.524 liters of water should be used.

Step-by-step solution

Calculate the moles of phosphorous acid needed

To find the number of moles of phosphorous acid needed, we can first find the mass of the required phosphorous acid and then convert it to moles using molar mass. The mass of phosphorous acid can be found using the given density and volume: Mass = Density × Volume $$ \text{Mass of }\mathrm{H}_{3} \mathrm{PO}_{3} = 1.651 \frac{\mathrm{g}}{\mathrm{cm}^{3}} \times 0.250 L \times \frac{1000 \mathrm{cm}^{3}}{1 L} = 412.75 g $$ Now, we can calculate the moles of \(\mathrm{H}_{3} \mathrm{PO}_{3}\) using its molar mass (81.995 g/mol): $$ \text{Moles of }\mathrm{H}_{3} \mathrm{PO}_{3} = \frac{412.75 g}{81.995 \frac{\mathrm{g}}{\mathrm{mol}}} = 5.035 \text{ moles} $$

Calculate the moles of phosphorus triiodide needed

To calculate the moles of phosphorus triiodide needed, we must consider that the overall yield of the reaction is 75%. This means that to obtain the desired 5.035 moles of phosphorous acid, we will need 1.333 times more moles of phosphorus triiodide: $$ \text{Moles of }\mathrm{PI}_{3} = 5.035 \text{ moles } \times \frac{1}{0.750} = 6.713 \text{ moles} $$

Calculate the grams of phosphorus triiodide needed

Using the molar mass of \(\mathrm{PI}_{3}\) (411.687 g/mol), we can now find out how many grams of phosphorus triiodide should be weighed out: $$ \text{Mass of }\mathrm{PI}_{3} = 6.713 \text{ moles} \times 411.687 \frac{\mathrm{g}}{\mathrm{mol}}= 2763.53 g $$

Calculate the moles of water needed

According to the stoichiometry of the balanced equation, for every mole of phosphorus triiodide, 3 moles of water are needed. However, we must also consider the 45% excess of water in the process: $$ \text{Moles of }\mathrm{H}_{2}\mathrm{O} = 6.713 \text{ moles} \times 3 \times 1.450 = 29.148 \text{ moles} $$

Calculate the volume of water needed

Finally, using the given density of water (\(1.00 \mathrm{~g} / \mathrm{cm}^{3}\)), we can find out what volume of water should be used: $$ \text{Volume of }\mathrm{H}_2 \mathrm{O} = \frac{29.148 \text{ moles} \times 18.015 \frac{\mathrm{g}}{\mathrm{mol}}}{1.00 \frac{\mathrm{g}}{\mathrm{cm}^{3}}} \times \frac{1 \mathrm{mL}}{1 \mathrm{cm}^{3}} \times \frac{1L}{1000 \text{mL}}= 0.524L $$ So, 2763.53 grams of phosphorus triiodide and 0.524 liters of water should be used to obtain 0.250 liters of phosphorous acid.

Key concepts

Mole-to-Mole Ratios in Chemical Reactions

Understanding mole-to-mole ratios is crucial when balancing chemical equations and predicting the amount of reactants needed and products formed. Let's dive into what these ratios tell us using our textbook problem as an example.

Mole-to-mole ratios are derived from the coefficients in a balanced chemical equation. For instance, the reaction between phosphorus triiodide (\textbf{PI}\(_3\)) and water (\textbf{H}\(_2\)O) to form phosphorous acid (\textbf{H}\(_3\)PO\(_3\)) and hydrogen iodide (\textbf{HI}) can be represented as:\[\mathrm{PI}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(s)+3 \mathrm{HI}(g)\]From this, we can see the mole-to-mole ratios: for every 1 mole of PI\(_3\), 3 moles of H\(_2\)O are required, and 1 mole of H\(_3\)PO\(_3\) and 3 moles of HI are produced. The coefficients in front of the chemical formulas indicate these ratios. In the student's exercise, the ratios were particularly useful in determining the amounts of PI\(_3\) and water needed for the desired amount of phosphorous acid.But what if the reaction doesn't go to completion or isn't perfectly efficient? That's where the concept of reaction yield comes to play.

Reaction Yield Calculation

The reaction yield tells us how much product we can actually expect compared to the theoretical maximum set by the balanced equation. It's expressed as a percentage, and in our example, the student's reaction has a yield of 75%. The yield is critical in determining the real amounts of reactants needed.

To calculate the theoretical yield, we consider the complete reaction of the limiting reactant. In the exercise, if we get 5.035 moles of phosphorous acid from the complete reaction of PI\(_3\), this is the theoretical yield. However, with a practical yield of 75%, we only produce 75% of this amount. To accommodate the reality of the lower yield, we need more PI\(_3\) initially to end up with the desired product quantity (5.035 moles of phosphorous acid in this case).

You'll often need to adjust the amounts of reactants when calculating for the actual yield of the reaction. Simply, this means multiplying the theoretical mole quantity by the reciprocal of the percent yield (as a decimal) to find the actual moles of reactants needed:\[\text{Actual moles of reactant} = \text{Theoretical moles} \times \frac{1}{\text{Percent Yield}}\]By keeping this in mind, you're less likely to run short on your reactants!

Understanding Excess Reactant

In our exercise, there's an excess of water; 45% more than what the stoichiometry would suggest. An excess reactant is used, often intentionally, to ensure that the other reactant is completely consumed and helps drive the reaction to completion. Identifying and calculating for an excess reactant can greatly impact the success of an experiment.

For every mole of PI\(_3\), 3 moles of water are usually required, according to the mole-to-mole ratio. But, given the 45% excess, we actually need 145% of the stoichiometric amount of water. This is calculated by multiplying the stoichiometric moles of water by 1.45. By doing so, we ensure that all of the PI\(_3\) reacts.

Why is this important? In the laboratory, reactions may not proceed as expected due to various side reactions or incomplete interactions among reactants. Having an excess of one reactant can increase the likelihood that the other reactant is fully utilized, optimizing the yield. This tactic is commonly used when the excess reactant is cheaper or more readily available than the other substances involved.