Worked Solutions to Old Versions of Gas Problems

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John Gervase Gray
5 years ago
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1 Worked Solutions to Old Versions of Gas Problems 1. Urea (H 2NCONH 2) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide: 2NH 3(g) + CO 2(g) -----> H 2NCONH 2(s) + H 2O(g) Ammonia gas at 223 degrees C and 90. atm flows into the reactor at a rate of 500. L/minute. Carbon dioxide at 223 degrees C and 45 atm flows into the reactor at a rate of 600 L/minute. What mass of urea is produced per minute by this reaction assuming 100% yield? Solution: Plan, working backwards: We need grams of urea. To get that, we need moles of urea. To get moles of urea, we need moles of limiting reactant. To know which reactant is limiting, we need to know moles of both reactants, compared to the mole ratio in the equation. And to find moles of both reactants, we use the ideal gas law. This is a stoichiometry problem, so we should work in moles. We are given temperature, pressure, and volume (per minute) for each gas, and told to find the yield per minute. With T, P, and V, finding n isn't too bad. We just need to keep our units straight and rearrange PV = nrt. PV = nrt ----> divide both sides by RT -----> PV / RT = n Since our units will be L, atm, moles, and K, we should use the version of the ideal gas constant R that uses those units: R = L * atm / mol * K. Temperatures should be in K, so add to each. This gives: [(90 atm)(500 L/min)]/[( L atm/mol K)( K)] = moles of ammonia/min [(45 atm)(600 L/min)]/[( L atm/mol K)( K)] = moles of carbon dioxide/min. Of these two reagents, the limiting reagent will determine how much product we produce. According to the equation, we need twice as much ammonia as carbon dioxide - the mole ratio for ammonia to CO 2 is 2:1. However, looking at our actual ratio (1105:663), we see that it's less than 2:1. In other words, we are short on ammonia. Ammonia is limiting. [Another way to find the limiting reagent is to solve for both reagents as limiting, and then choose the smaller yield. This is time-consuming.] Since ammonia is limiting, the number of moles of ammonia will determine the number of moles of urea product. The ratio of ammonia to urea is 1:2, according to the equation, so there should be half as many moles of urea as moles of ammonia: moles ammonia/min x = moles of urea

2 To convert moles of urea to grams, we first find the molar mass of urea. Urea (H 2NCONH 2) has 4 hydrogens, 1 carbon, 1 oxygen, and 2 nitrogens. This gives a molar mass of 4* *14.01 or about g/mol. Converting moles to grams: moles urea x g/mol = grams urea. 2. Methanol, CH 3OH, can be produced by the following reaction: CO(g) + 2H 2(g) ---> CH 3OH (g) Hydrogen at STP (standard temperature and pressure; look this up if you don't know it!) flows into a reactor at a rate of 16.0 L/minute. Carbon monoxide at STP flows into a reactor at a rate of 25.0 L/min. If 5.30 grams of methanol is produced per minute, what is the percent yield for this reaction? Solution: Plan, working backwards: To get percent yield, compare actual yield to expected yield. To get expected yield in grams, we need to find expected yield in moles. To get yield in moles, find moles of limiting reactant and use the mole ratio. To get moles of limiting reactant, use PV = nrt to find moles of both reactants and compare their mole ratio to the balanced equation. This is almost the same as the last problem, with the addition of a final step to get percent yield. STP = 1 atm, 0 C ( K). (NOTE: Some people use atm, or 100,000 Pa as STP.) Moles of hydrogen = PV / RT = [(1 atm)(16.0 L/min)]/[( L atm/mol K)( K)] =.714 moles/min hydrogen gas There is also a shortcut! At STP, one mole of an ideal gas will occupy ~ L. So: 16.0 L hydrogen/min x (1 mole / L) =.714 moles hydrogen gas / min We'll use that shortcut to get moles of carbon monoxide: 25.0 L CO/min x (1 mole / L) = moles carbon monoxide/min This reaction requires a 2:1 ratio of H 2 to CO, and yet we have less H 2. H 2 is limiting. So we'll use the moles of H 2 and the mole ratio of the balanced equation to find moles of methanol..714 moles H 2 x =.357 moles methanol. The MW of methanol is ( * * ) g/mol = g/mol

3 Our expected yield in grams is therefore:.357 moles x g/mol = grams. Our actual yield was 5.30 grams. We divide actual yield by expected yield and multiply by 100% to get percent yield: 3. A 2.00 L sample of O 2(g) was collected over water at a TOTAL pressure (with the water vapor included) of 785 torr at 25 degrees Celsius. When this O 2 was dried (removing the water vapor), the gas had a volume of 1.94 L at 25 degrees Celsius and 785 torr. Calculate the vapor pressure of water at 25 degrees Celsius. A relatively straightforward plan, working backwards: I think the hint I gave here was less helpful than desired - it requires a leap of logic to take a shortcut! The simplest way to get to the answer without confusion is this: * To get vapor pressure of water in the oxygen/water mixture at 25 C, we can take the total pressure of 785 torr and subtract the partial pressure of oxygen. The remainder must come from the water vapor. * To find the partial pressure of oxygen in that oxygen/water mixture, we need moles, volume, and temperature for oxygen for that oxygen/water mixture. * We know that temperature is 25 C ( K) and volume is 2.00 L (for the oxygen/water mixture), so we just need moles of oxygen in that oxygen/water mixture. * Drying out the water didn't remove any oxygen. So the moles of oxygen at the second time point (dried pure oxygen) must be the same as the moles of oxygen at the first time point (oxygen/water mixture). We can find the moles of oxygen once the water has been dried using PV/RT. A good general rule for dealing with problems that involve changes between 2 time points is to work out what is CONSTANT between those time points, as well as what is changing! That gives you a number to start from! So the solution path going forward is: Pressure, volume, and temperature of pure O 2 -> moles O 2 -> Pressure O 2 in mixture -> Pressure of water by subtraction of pressure of oxygen First, we convert all pressure to atm and all temperature to K: 25 C = K 785 torr x = atm Then we find moles of oxygen:

4 n = PV/RT = [(1.033 atm)(1.94 L)]/[( L atm/mol K)*( K)] =.0819 moles O 2 Then we find the partial pressure of O 2 in the original oxygen/water mix. P = nrt/v = [(.0819 moles)( L atm/mol K)( K)]/(2.00 L) = atm atm x (760 torr / 1 atm) = torr So the partial pressure of oxygen was torr. The total pressure was 785 torr. If we subtract, the partial pressure of oxygen must be 23.6 torr. We can check this by looking up the vapor pressure of water at 25 C online, and we find a result of 23.8 torr; this is within experimental error. AN ALTERNATIVE SOLUTION THAT REQUIRES A LEAP OF INSIGHT, UNDERSTANDING OF DALTON'S LAW OF PARTIAL PRESSURES, AND EXACTLY ONE STEP: There is ANOTHER path to the solution, the shortcut I hinted at, but it's more subtle: The total pressure of the oxygen/water mixture is 785 torr. It consists of 2 L of oxygen/water. When we remove the water, the volume decreases to 1.94 L. Because ideal gases are treated as identical to each other, we can say the original mix is.06 parts water to 1.94 parts oxygen - in other words, each gas's contribution is proportional to its partial pressure in the original mixture. This is a risky approach because it's easy to get confused by this kind of thinking! Adding, say, 1 L of oxygen and 1 L of water vapor does NOT necessarily give you 2 L of mixture, since gases don't have fixed volume. But it does work in this case: 4. A 1.42 gram sample of a pure compound, with formula M 2SO 4 [M is some unknown metal cation], was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in precipitation of all the sulfate ions as calcium sulfate. The precipitate was dried and found to weigh 1.36 grams. Determine the atomic mass of M and identify M. Note: There was a major error in my hint for this problem! I wrote MCl 2 for some strange reason, not MCl. I apologize for any resulting confusion. The mole ratio of CaSO 4 to M 2SO 4 remains unchanged at 1:1. M 2SO 4 + CaCl 2 -> 2MCl + CaSO 4 Solution plan (working backwards). To identify M, we can compare M's atomic mass to elements on the periodic table. To find M's atomic mass, we can find the molar mass of M 2SO 4, subtract the mass of SO 4 to get the mass of M 2, and then divide by 2.

5 To find the molar mass of M 2SO 4, we can divide the number of grams of M 2SO 4 (given) by the number of moles of M 2SO 4. To find the number of moles of M 2SO 4, we can find the number of moles of CaSO 4 precipitate, then use the balanced reaction to get the mole ratio of M 2SO 4 to CaSO 4. To find moles of CaSO 4 precipitate, we can divide the grams of CaSO 4 by the molar mass of CaSO 4. Find molecular weight of CaSO 4: MW CaSO 4 = (calcium) (sulfur) + 4*16.0 (oxygen) g/mole = g/mol Use MW to get moles of CaSO 4 precipitate: 1.36 g CaSO 4 x =.0100 moles CaSO 4 Mole ratio of M 2SO 4 to CaSO 4 in balanced equation is 1:1: =.0100 moles M 2SO 4 We know the mass of M 2SO 4 present. Since MW = mass/moles, we can calculate MW: MW M 2SO 4 = 1.42 g M 2SO 4 /.0100 moles M 2SO 4 = 142 g/mole We subtract the molar mass of SO 4 to get the molar mass of M 2: 142 g/mol - ( *16) g/mol = g/mol M 2 We divide molar mass of M 2 by 2 to get molar mass of M: g/mol M 2 / 2 = g/mol M Looking at the periodic table, M is most likely sodium, which has MW g/mol. This makes sense. If the compound is M 2SO 4, and sulfate ions have a -2 charge, M ions must have a +1 charge. So our compound was Na 2SO 4.

c. K 2 CO 3 d. (NH 4 ) 2 SO 4 Answer c

c. K 2 CO 3 d. (NH 4 ) 2 SO 4 Answer c Chem 130 Name Exam 2, Ch 4-6 July 7, 2016 100 Points Please follow the instructions for each section of the exam. Show your work on all mathematical problems. Provide answers with the correct units and