Sample Exam 3 - roberts(eer474 – Sample Exam Three –...

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Unformatted text preview: roberts (eer474) – Sample Exam Three – Shear – (52375) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. DrRuth says: In all cases, select the answer that BEST answers the question asked. Read the whole thing first and plan which to answer first because you feel confident about them. Notice the harder ones and save them up till last so you don’t use up all your time working on them. Shear 52375 g = 9 . 8 m / s 2 1 atm = 101325 Pa = 760 torr R = 0.08206 L atm K − 1 mol − 1 R = 0.08314 L bar K − 1 mol − 1 R = 8.314 J K − 1 mol − 1 ( P + a ( n/V ) 2 )( V- nb ) = nRT 001 10.0 points Consider the reaction 2 CH 4 (g) + 3 O 2 (g) + 2 NH 3 (g) → 2 HCN(g) + 6 H 2 O(g) If 128 g each of NH 3 and CH 4 and an excess of O 2 are reacted, what mass of HCN can be produced? 1. 203 g HCN correct 2. 610 g HCN 3. The answer cannot be determined from information available. 4. 419 g HCN 5. 216 g HCN Explanation: m NH 3 = 128 g m CH 4 = 128 g We recognize this as a limiting reactant problem because the amounts of more than one reactant are given. We must determine which of these would be used up first (the limiting reactant). We calculate the number of moles of each reactant available : ? mol NH 3 = 128 g NH 3 × 1 mol NH 3 17 g NH 3 = 7 . 53 mol NH 3 ? mol CH 4 = 128 g CH 4 × 1 mol CH 4 16 g CH 4 = 8 mol CH 4 and then compare the required ratio of reac- tants to the available ratio of reactants. The balanced chemical equation shows that we need 2 mol CH 4 for every 2 mol of NH 3 . We use these coefficients to calculate the required ratio of reactants: 2 mol CH 4 2 mol NH 3 = 1 mol CH 4 1 mol NH 3 From this ratio we see that each mole of NH 3 that reacts requires exactly 1 mole of CH 4 . Next we calculate the available ratio of reactants from our data: 8 mol CH 4 7 . 53 mol NH 3 = 1 . 06 mol CH 4 1 mol NH 3 We see that we have 1.06 moles of CH 4 available for each mole of NH 3 , so we have more than enough CH 4 to react with all of the NH 3 . We will run out of NH 3 first, so NH 3 is the limiting reactant. We use the amount of NH 3 as the basis for further calculations. We use the mole ratio from the chemical equation and the molar mass of HCN to calculate grams HCN produced. ? g HCN = 7 . 53 mol NH 3 × 2 mol HCN 2 mol NH 3 × 27 g HCN 1 mol HCN = 203 . 31 g HCN 002 10.0 points Write the net ionic equation for this reac- tion occuring in water: Barium perchlorate and sodium sulfate are mixed to form sodium perchlorate and barium sulfate. roberts (eer474) – Sample Exam Three – Shear – (52375) 2 1. 2 Na + + 2 ClO − 3 → 2 NaClO 3 2. SO 2 − 4 + Ba(ClO 4 ) 2 → 2 ClO − 4 + BaSO 4 3. 2 Na + + 2 ClO − 4 → 2 NaClO 4 4. SO 2 − 4 + Ba 2+ → BaSO 4 correct 5. No reaction occurs....
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Sample Exam 3 - roberts(eer474 – Sample Exam Three –...

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