Ryan Lenet MISE - Physical Basis of Chemistry Second Set of Problems - Due November 13, 2005

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1 Ryan Lenet MISE - Physical Basis of Chemistry Second Set of Problems - Due November 13, 2005 Page 1 Submit electronically (digital drop box) by Sunday, November 13 by 6 pm. Note: When submitting to digital Drop Box label your files with your name first and then the label - HmwkTwo. Please put you name in the Header along with the already-inserted page #. There is a useful equation template document posted below this assignment in Bb. Give it a look. The following information may be useful when solving the problems: Assume that all gases behave ideally, i.e., PV = nrt ; where R = L atm mole -1 K mole = x things. The atomic weight of an element listed in the periodic table is the mass of one mole of atoms in grams. The molecular weight - determined from molecular formula as the sum of the atomic weights of all of the constituent atoms - is the mass of one mole of molecules in grams. 1 atmosphere (atm) = 760 mm Hg = 76 cm Hg = kilopascals grams = 1 lb ; 1000 cm 3 = 1 L ; 2.54 cm = 1 inch ; {32ºF = 0ºC & 212ºF = 100ºC} In case you have to figure out the linear relationship to convert ºF <----> ºC. K = ºC Volume of a cylinder = (radius) 2 (height) 1. A child is holding two balloons A and B each containing a different gas on a day when the local weatherman reports the barometric pressure to be 29 inches of mercury and a temperature of 78ºF. If balloon A contains 6.00 L of nitrous oxide (N 2 O(g)) and balloon B contains 8.00 L of helium (He(g)), determine the moles and grams of each gas in each balloon. 1. A L ( N2O(g)) B L (He (g)) A. 29 inches = 2.54cm/1inch = cm Hg cm Hg 10 mm/1cm = mm Hg mmhg = 1 atm/760mmhg =.9692 atm

2 78 º F = 5/9(78-32) = ºC K= º C K = º K Page 2 PV=nRT n = PV/RT (.9692 atm)(6.00 L)/(,0821 L atm molesk)( ºK) n = / 24.51moles =.2373 moles of N2O A= (44 g/mole (MW) (.2373 moles) = g of N2O B. n = (.9692 atm)(8.00 L)/ (.0821 L atm molesk)( ºK) n= / moles =.3163 moles of He B = (4 g/mole) (.3173 moles) = g of He 2. An ideal gas at 1.00 atmosphere was collected in a bulb of unknown volume V. A valve is opened which allows the gas to expand into a previously evacuated bulb whose volume is known to be exactly L. When things settle down (the gas fills the full volume of both bulbs), it is noted that the temperature had not changed, and the gas pressure was 530 mm Hg. What is the volume, V, of the first bulb (in liters)? 2. 1 atm V=? volume of bulb =.500 L P= 530 mmhg PV= nrt 530mmHg (1atm/760mmHg) =.6974atm PiVi/Ti = PfVf/Tf Vf= Vi(Pi/Pf)(Tf/Ti) = Vf=Vi(Pi/Pf) Vf = (.500L) ( 1 atm/.6974) =.7169 L 3. Balancing a chemical reaction of gases and reasoning with moles Gaseous propane (C 3 H 8 (g)) is a hydrocarbon that is often used as the fuel for small gas grills. The burning of propane is actually the reaction with molecular diatomic oxygen (O 2 (g)) a combustion to produce carbon dioxide (CO 2 ) and water (steam): C 3 H 8 (g) + O 2 (g) > CO 2 (g) + H 2 O(g). 3. a. C3H8(g) + O2(g) =====> CO2(g) + H2O(g) C3H8 (g) + 5O2(g) =====> 3CO2(g) + 4H2O(g) Every one mole of C3H8 propane will react with 5 moles of O2 to yield or produce 3 moles of Carbon dioxide CO2 and 4 moles of H2O water. (a) According to Dalton, atoms must be conserved, i.e., there must be the same number

3 Page 3 of atoms of each element on both sides of the >. Please place integer coefficients in front of each chemical species as needed to conserve the atoms, i.e., balance the reaction. How should we interpret these coefficients? Read on. When balanced, the coefficients tell us the correct recipe. [For example, the balanced reaction: 2 H 2 + O > 2 H 2 O ; can be read as follows: Every 2 molecules of H 2 will react with one molecule of O 2 to yield or produce 2 molecules of H 2 O. Since the relationship between grams and moles is such a useful one (recall the first homework), we can always count the molecules in moles (similar to counting in dozens). Then, we can interpret the coefficients above as follows: Every 2 moles of H 2 will react with one mole of O 2 to yield or produce 2 moles of H 2 O. This will be useful in (c) below.] 3. a. C3H8(g) + O2(g) =====> CO2(g) + H2O(g) C3H8 (g) + 5O2(g) =====> 3CO2(g) + 4H2O(g) Every one mole of C3H8 propane will react with 5 moles of O2 to yield or produce 3 moles of Carbon dioxide CO2 and 4 moles of H2O water. (b) A tank of propane is advertised as containing 5 lbs. of propane and the tank - assumed to be a cylinder - has the listed dimensions of 8 in diameter x 12.5 in height. The temperature of the propane in the tank is 59ºF. Please determine the following. The mass of propane in the tank in grams. mass of propane tank 5 lbs. (16oz./1lb)(28.35g)(1 oz.) = 2,268 g Volume of cylinder - L x x r^ x 3.14 x 16 = 628 in.^3 The volume of propane in the tank in liters Volume of propane tank Volume of cylinder - L x x r^ x 3.14 x 16 = 628 in.^3 628 in.^3 ( cm^3 / 1in.^3) ( 1mL / 1cm^3) (1L/ 1000mL) /1000 = L The temperature of propane in the tank in ºC and in K. Temperature of propane tank 5/9(59-32) = 15ºC 15ºC = 288ºK The moles of propane in the tank. The moles of the propane tank C3H8 = = 44g/mole 2,268 g/44g/mole = moles The number of propane molecules in the tank. Molecules of propane moles C3H8 ( x 10^23 molecules C3H8 / 1mole C3H8) = x 10^25 molecules or 3.1 x 10^25 molecules of C3H8 The pressure of propane in the tank in atmospheres - assuming ideal gas behavior.

4 Pressure of propane in tank PV= nrt P= nrt/v Page 4 P= (51.55 moles)(0.0821latmmolesk)(288ºk)/ 10.29L) P= atm/10.29 = atm (c) If all of the propane in the tank was completely reacted with the exact proportion of O 2 (g) indicated in your balanced reaction, how many moles of O 2 (g) would be required? How many molecules of O 2 (g) is this? c. Moles of O2 2,268 g C3H8 ( 1moles C3H8/44 g C3H8) = mol mol O2 = mol C3H8 ( 5mol O2/ 1 molecule) = mol O mol O2 ( X 10^23/ 1 mol O2) = 1.55 x 10^26 molecules of O2 (d) How many moles of CO 2 (g) and how many moles of H 2 O(g) should be produced by the amount of C 3 H 8 (g) and O 2 (g) determined in (c)? d. mol CO2 = ( mol C3H8)( 3mol CO2/1 mol C3H8) = mol CO2 mol H2O = ( mol C3H8)( 4 mol H2O/1 mol C3H8) = mol of H2O (e) Extra Credit: Air is a mixture of gases and is popularly proclaimed to contain 20 % oxygen (O 2 (g)) and 80 % nitrogen (N 2 (g)). This percentage is a mole percent - not a mass percent. Thus, every 100 moles of air contains 20 moles of O 2 (g). How many moles of air would contain the amount of O 2 (g) determined in (c)? 4. An allotrope is one of several forms of the same element with different structure. Consider the element carbon (C). Several years ago an allotrope of carbon - neither graphite nor diamond - was discovered and named buckminsterfullerine or buckyball, C X, where X is a whole number. A sample of C X (g) at o C and mm Hg is found to have a density of g/l. Determine X and hence the formula C X for one of these buckyball molecules. 4. Cx where x is a whole number Cx(g) at ºC and 680 mmhg with a density of 11.67g/L RTd/P= MW (.0821LatmmolK)(673ºK)(11.67g/L) / (.8947 atm) / 8947= g/mol 720.7g/mol of C ( 1 mol C/ 12 g ) = 60 mol

5 x= 60 C60 Page 5