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Chapter 3: Problem 80

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?

Short Answer

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

Step by step solution

01

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.
02

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 \(1(3)+15.999(3.003)+30.974 = 81.995 \frac{g}{mol}\).
03

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%).
04

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 \(30.97 + 126.9(3) = 411.67 \frac{g}{mol}\).
05

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.
06

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.

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Most popular questions from this chapter

Chromium (atomic mass \(=51.9961\) amu) has four isotopes. Their masses are \(49.94605 \mathrm{amu}, 51.94051 \mathrm{amu}, 52.94065 \mathrm{amu}\), and \(53.93888\) amu. The first two isotopes have a total abundance of \(87.87 \%\), and the last isotope has an abundance of \(2.365 \%\). What is the abundance of the third isotope? Estimate the abundances of the first two isotopes.

When \(4.0 \mathrm{~mol}\) of \(\mathrm{CCl}_{4}\) reacts with an excess of \(\mathrm{HF}, 3.0 \mathrm{~mol}\) of \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) (Freon) is obtained. The equation for the reaction is $$ \mathrm{CCl}_{4}(l)+2 \mathrm{HF}(g) \longrightarrow \mathrm{CCl}_{2} \mathrm{~F}_{2}(l)+2 \mathrm{HCl}(g) $$ State which of the statements are true about the reaction and make the false statements true. (a) The theoretical yield for \(\mathrm{CCl}_{2} \mathrm{~F}_{2}\) is \(3.0 \mathrm{~mol}\). (b) The theoretical yield for \(\mathrm{HCl}\) is \(71 \mathrm{~g}\). (c) The percent yield for the reaction is \(75 \%\). (d) The theoretical yield cannot be determined unless the exact amount of \(\mathrm{HF}\) is given. (e) From just the information given above, it is impossible to calculate how much HF is unreacted. (f) For this reaction, as well as for any other reaction, the total number of moles of reactants is equal to the total number of moles of product. (g) Half a mole of \(\mathrm{HF}\) is consumed for every mole of \(\mathrm{CCl}_{4}\) used. (h) At the end of the reaction, no \(\mathrm{CCl}_{4}\) is theoretically left unreacted.

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