Stoich02answers

# Stoich02answers - Cheung Anthony Stoich 2 Due midnight Inst...

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Cheung, Anthony – Stoich 2 – Due: Oct 29 2006, midnight – Inst: McCord 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Balance the equation ? H 2 CO + ? H 2 O ? CH 4 + ? O 3 . The coe±cients are 1. 3; 2; 1; 2. 2. 3; 3; 3; 2. correct 3. 2; 1; 2; 1. 4. 3; 3; 2; 2. 5. 1; 2; 2; 3. Explanation: A balanced equation has the same num- ber oF each kind oF atom on both sides oF the equation. We fnd the number oF each kind oF atom using equation coe±cients and com- position stoichiometry. ²or example, we fnd there are 12 H atoms on the product side: ? H atoms = 3 CH 4 × 4 H 1 CH 4 = 12 H The balanced equation is 3 H 2 CO + 3 H 2 O 3 CH 4 + 2 O 3 , and has 3 C, 12 H and 6 O atoms on each side. The equation coe±cients are 3, 3, 3, 2. 002 (part 1 oF 1) 10 points Consider the reaction N 2 + 3H 2 2 NH 3 . How much NH 3 can be produced From the reaction oF 74.2 g oF N 2 and 14.0 moles oF H 2 ? 1. 5 . 62 × 10 24 molecules 2. 1 . 26 × 10 25 molecules 3. 1 . 59 × 10 24 molecules 4. 3 . 19 × 10 24 molecules correct 5. 1 . 69 × 10 25 molecules Explanation: m N 2 = 74.2 g n H 2 = 14.0 mol ²irst you must determine the limiting reac- tant: ? mol N 2 = 74 . 2 g N 2 × 1 mol N 2 28 g N 2 = 2 . 65 mol N 2 According to balanced equation, we need 3 mol H 2 1 mol N 2 . We have 14 . 0 mol H 2 2 . 65 mol N 2 = 5 . 28 mol H 2 1 mol N 2 ThereFore, H 2 is an excess and N 2 is limiting. ? molec NH 3 = 2 . 65 mol N 2 × 2 mol NH 3 1 mol N 2 × 6 . 022 × 10 23 NH 3 molec 1 mol NH 3 molec = 3 . 19 × 10 24 molec NH 3 003 (part 1 oF 1) 10 points Consider the reaction 4 ²e(s) + 3 O 2 (g) 2 ²e 2 O 3 (s). IF 12.5 g oF iron(III) oxide (rust) are pro- duced From 8.74 g oF iron, how much oxygen gas is needed For this reaction? 1. None oF these correct 2. 8.74 g 3. 21.2 g 4. 7.5 g 5. 12.5 g

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Cheung, Anthony – Stoich 2 – Due: Oct 29 2006, midnight – Inst: McCord 2 Explanation: m iron = 8.74 g m oxide = 12.5 g The balanced equation for the reaction tells us that 4 mol Fe reacts with 3 mol O 2 to produce 2 mol Fe 2 O 3 . We have two possible starting points. We know 12.5 g Fe 2 O 3 was produced and that 8.74 g Fe was present at the start of the reaction. Choosing the 12.5 g of Fe 2 O 3 to start with, ±rst we convert to moles using the molar mass: ? mol Fe 2 O 3 = 12 . 5 g Fe 2 O 3 × 1 mol Fe 2 O 3 159 . 7 g Fe 2 O 3 = 0 . 0783 mol Fe 2 O 3 Now we use the mole ratio from the bal- anced equation to ±nd moles O 2 needed to produce 0.0783 mol Fe 2 O 3 . ? mol O 2 = 0 . 0783 mol Fe 2 O 3 × 3 mol O 2 2 mol Fe 2 O 3 = 0 . 117 mol O 2 We convert from moles to grams: ? g O 2 = 0 . 117 mol O 2 × 32 g O 2 1 mol O 2 = 3 . 744 g O 2 Starting with 8.74 g Fe and following the same steps results in the same numerical an- swer. 004
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Stoich02answers - Cheung Anthony Stoich 2 Due midnight Inst...

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