Problem 80 - Chemistry: Principles and Reactions

Question

A student prepares phosphorous acid, \(\mathrm{H}_{3} \mathrm{PO}_{3}\), by reacting solid phosphorus triodide 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 \(\mathrm{H}_{3} \mathrm{PO}_{3}\left(d=1.651 \mathrm{~g} / \mathrm{cm}^{3}\right)\). 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?

Step-by-Step Solution

Answer

Answer: After following the step-by-step calculations, the mass of phosphorus triiodide required and the volume of water needed can be obtained.

Step 1: Calculate the Mass of Phosphorous Acid Required

To obtain the mass of the desired product, phosphorous acid (H3PO3), multiply its density by the desired volume. $$ m_{H_{3}PO_{3}} = d\times V $$ Where \(m_{H_{3}PO_{3}}\) is the mass of phosphorous acid, \(d =1.651 \frac{g}{cm^3}\) is the density of phosphorous acid, and \(V = 0.250 L\) is the desired volume of phosphorous acid.

Step 2: Calculate the Moles of Phosphorous Acid Required

Convert the mass of phosphorous acid to moles using the molar mass of phosphorous acid (\(H_{3}PO_{3}\)). $$ n_{H_{3}PO_{3}} = \frac{m_{H_{3}PO_{3}}}{M_{H_{3}PO_{3}}} $$ Where \(n_{H_{3}PO_{3}}\) is the moles of phosphorous acid, \(m_{H_{3}PO_{3}}\) is the calculated mass of phosphorous acid from Step 1, and \(M_{H_{3}PO_{3}}\) is the molar mass of phosphorous acid, which can be calculated as \(1(3)+15.999(3.003)+30.974 = 81.995 \frac{g}{mol}\).

Step 3: Calculate the Moles of Phosphorus Triiodide Required

Based on the stoichiometry of the reaction, one mole of phosphorus triiodide \((PI_{3})\) produces one mole of phosphorous acid \((H_{3}PO_{3})\). However, we have to consider the reaction yield of 75%. Calculate the moles of phosphorus triiodide needed as follows: $$ n_{PI_{3}} = \frac{n_{H_{3}PO_{3}}}{0.75} $$ Where \(n_{PI_{3}}\) is the moles of phosphorus triiodide, \(n_{H_{3}PO_{3}}\) is the moles of phosphorous acid from Step 2, and 0.75 is the given reaction yield (75%).

Step 4: Calculate the Mass of Phosphorus Triiodide Required

Convert the moles of phosphorus triiodide to mass using the molar mass of phosphorus triiodide (\(PI_{3}\)). $$ m_{PI_{3}} = n_{PI_{3}}\times M_{PI_{3}} $$ Where \(m_{PI_{3}}\) is the mass of phosphorus triiodide, \(n_{PI_{3}}\) is the moles of phosphorus triiodide from Step 3 and \(M_{PI_{3}}\) is the molar mass of phosphorus triiodide, which can be calculated as \(30.97 + 126.9(3) = 411.67 \frac{g}{mol}\).

Step 5: Calculate the Moles of Water Required

Based on the stoichiometry of the reaction, 1 mole of phosphorus triiodide (\(PI_{3}\)) requires 3 moles of water (\(H_{2} O\)). Calculate the moles of water needed by considering the 45% excess, as follows: $$ n_{H_{2} O} = n_{PI_{3}}\times \frac{3}{1}\times 1.45 $$ Where \(n_{H_{2} O}\) is the moles of water, \(n_{PI_{3}}\) is the moles of phosphorus triiodide from Step 3, the ratio of \(3:1\) comes from the stoichiometry of the balanced equation given, and 1.45 accounts for the 45% excess of water required.

Step 6: Calculate the Volume of Water Required

Convert the moles of water to volume using the density of water (\(1.00 \frac{g}{cm^3}\)) and the molar mass of water \((H_{2} O)\) as follows: $$ V_{H_{2} O} = \frac{n_{H_{2} O}\times M_{H_{2} O}}{d_{H_{2} O}} $$ Where \(V_{H_{2} O}\) is the volume of water, \(n_{H_{2} O}\) is the moles of water from Step 5, \(M_{H_{2} O} = 1(2) + 15.999 = 18.015 \frac{g}{mol}\) is the molar mass of water, and \(d_{H_{2} O} = 1.00 \frac{g}{cm^3}\) is the density of water. Upon performing these calculations, you will obtain the mass of phosphorus triiodide to be weighed out and the volume of water to be used.

Key Concepts

Phosphorus Triiodide

Phosphorus triiodide, represented chemically as \( \text{PI}_3 \), is a covalent compound composed of one phosphorus atom and three iodine atoms. This compound is commonly used as a reagent in the production of phosphorous acid from water.

Despite being typically solid at room temperature, phosphorus triiodide is sensitive to moisture and reacts vigorously with water to form phosphorous acid and hydrogen iodide gas.

When dealing with phosphorus triiodide, it’s crucial to handle it in a controlled environment to prevent premature reactions with water or other materials.

Phosphorous Acid

Phosphorous acid, with the chemical formula \( \text{H}_3\text{PO}_3 \), is also known by its assigned name dihydroxyphosphine oxide. In this process, it’s created as a product when phosphorus triiodide reacts with water.

This acid must be carefully monitored in chemical reactions due to its reactivity and the need for specific conditions to navigate its synthesis successfully.

In the biochemical and industrial chemical world, phosphorous acid is used for various purposes, including acting as a reagent in chemical syntheses and serving certain applications in agriculture as a fungicide.

Reaction Yield

Calculating the reaction yield is essential to understand how efficient a chemical reaction is. Reaction yield is the actual amount of product obtained from a reaction compared to the maximum amount possible (theoretical yield).

Here, 75% is noted as the yield for the production of phosphorous acid from phosphorus triiodide, indicating that only 75% of the theoretical amount is expected to form under practical conditions.

Yield efficiency can vary due to factors such as reactant purity, reaction conditions, and procedural losses. Thus, understanding the yield helps chemists plan for excess reactants to ensure sufficient product formation.

Excess Reactant Calculation

Determining the excess reactant is a crucial step to ensure that the reaction goes to completion, especially in this reaction involving phosphorus triiodide and water.

An excess reactant is used to drive a reaction to completion by providing more than the necessary stoichiometric amount. The instruction to use a 45% excess of water ensures that there is more than enough water available for the reaction with the phosphorus triiodide.

By calculating the excess reactant, chemists can avoid scenarios where a limiting reactant stops the reaction prematurely, potentially affecting the reaction yield and overall efficiency.