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Unformatted text preview: Version 357 – Exam 3 – Sutcliffe – (53770) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. THIS EXAM IS FOR Dr. SUTCLIFFE’s TTH SECTION ONLY. CHECK YOU BUB BLED YOUR NAME, EID AND VERSION NUMBER CORRECTLY. 001 10.0 points How many moles of oxygen atoms (not molecules) are present in 4 . 55 × 10 25 molecules of dinitrogen pentoxide (N 2 O 5 )? 1. 23 × 10 25 mol 2. 75.6 mol 3. 378 mol correct 4. 189 mol Explanation: n = 4 . 55 × 10 25 molec Each N 2 O 5 molecule contains five oxygen atoms. We can use Avogadro’s number and the ratio of O atoms to N 2 O 5 molecules to find the number of oxygen atoms: ? mol O = 4 . 55 × 10 25 molec N 2 O 5 × 5 atoms O 1 molec N 2 O 5 × 1 mol O 6 . 022 × 10 23 atoms O = 378 mol O 002 10.0 points 60.0 g O 2 and 50.0 g S are reacted according to the equation 2 S + 3 O 2 → 2 SO 3 . Which reactant is in excess and by how many grams? 1. O 2 ; 20.0 g 2. S; 24.8 g 3. O 2 ; 10.0 g 4. O 2 ; 24.8 g 5. S; 20.0 g 6. S; 10.0 g correct Explanation: m O 2 = 60 . 0 g m S = 50 . 0 g From the balanced equation we see that we need 3 mol O 2 2 mol S = 1 . 5 mol O 2 1 mol S From this ratio we see that each mole of S that reacts requires exactly 1.5 moles of O 2 . Next we need to determine how many moles of each reactant we actually have: ? mol O 2 = 60 . 0 g O 2 × 1 mol O 2 32 g O 2 = 1 . 875 mol O 2 ? mol S = 50 . 0 g S × 1 mol S 32 g S = 1 . 56 mol S Now we calculate the available ratio of reac tants and compare that to what is required: 1 . 875 mol O 2 1 . 56 mol S = 1 . 2 mol O 2 1 mol S We have 1.2 mol of O 2 for every mole of S. This is less than the 1.5 mol required, so O 2 is the limiting reactant and S is in excess. To determine by how much S is in excess, we need to calculate how many moles of S will react with the 60.0 g or 1.875 mol O 2 : ? mol S = 1 . 875 mol O 2 × 2 mol S 3 mol O 2 × 32 g S 1 mol S = 40 . 0 g S Finally we subract the amount of S that reacts from the amount of S that we started with to determine how much is left over: Version 357 – Exam 3 – Sutcliffe – (53770) 2 50 . 0 g S 40 . 0 g S = 10 . 0 g S 40.0 grams of S will react leaving 10.0 grams S unreacted. 003 10.0 points A(n) 0.65 mole quantity of O 2 originally at 85 ◦ C is cooled such that it now occupies 4.0 L at 20. ◦ C. What is the final pressure exerted by the gas? 1. 3.9 atm correct 2. Not enough information is given. 3. 6.2 atm 4. 0.27 atm 5. 0.92 atm 6. 1.1 atm Explanation: T = 20 ◦ C + 273 = 293 K V = 4 . 0 L n = 0 . 65 mol Using the Ideal Gas Law, P V = n RT P = n RT V = (0 . 65 mol) (0 . 082058 L · atm mol · K ) (293 K) 4 . 0 L = 3 . 9 atm 004 10.0 points Write the formula unit equation for this re action occurring in water: Potassium sulfate and barium chloride are mixed to form potas sium chloride and barium sulfate....
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This note was uploaded on 09/09/2009 for the course CH 302 taught by Professor Mccord during the Spring '09 term at University of TexasTyler.
 Spring '09
 MCCORD

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