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CH304K_HW06

# CH304K_HW06 - hyun(hh7953 Homework 6 Sparks(53435 This...

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hyun (hh7953) – Homework 6 – Sparks – (53435) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points If the reaction of 1.00 mole NH 3 (g) and 1.00 mole O 2 (g) 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O( ) is carried out to completion, 1. all of the NH 3 is consumed and 4.0 mol of NO(g) is produced. 2. all of the O 2 is consumed and 1.5 mol of NO(g) is produced. 3. all of the O 2 is consumed. correct 4. all of the O 2 is consumed and 4.0 mol of NO(g) is produecd. 5. all of the NH 3 is consumed. 6. all of the O 2 is consumed and 1.5 mol of H 2 O is produced. 7. 1.5 mol of H 2 O is produced. 8. all of the NH 3 is consumed and 1.5 mol of H 2 O is produced. 9. None of the other answers is correct. 10. 4.0 mol of NO(g) is produced and 1.5 mol of H 2 O is produced. Explanation: n NH 3 = 1 . 0 mol n O 2 = 1 . 0 mol 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). To do this we compare the required ratio of reactants to the available ratio of reactants. The balanced chemical equation shows that we need 4 mol NH 3 for every 5 mol O 2 . We use these coefficients to calculate the required ratio of reactants: 4 mol NH 3 5 mol O 2 = 0 . 8 mol NH 3 1 mol O 2 From this ratio we see that each mole of O 2 thatreacts requires exactly 0.8 mol NH 3 . Next we calculate the available ratio of reactants from our data: 1 mol NH 3 1 mol O 2 We have 1 mol of NH 3 available for each mole of O 2 , so we have more than enough NH 3 to react all of the O 2 . We will run out of O 2 first, so O 2 is the limiting reactant. Therefore it is true that all of the O 2 will be consumed in the reaction. We can calculate the amounts of NO and H 2 O that would be produced to check that the other answers are incorrect. Since the reaction must stop once we run out of O 2 , we use the amount of O 2 as the basis for these calculations. We use the mole ratios from the chemical equation to calculate moles NO and H 2 O produced: ? mol NO = 1 . 0 mol O 2 × 4 mol NO 5 mol O 2 = 0 . 80 mol NO ? mol H 2 O = 1 . 0 mol O 2 × 6 mol H 2 O 5 mol O 2 = 1 . 2 mol H 2 O 002 10.0 points Calculate the reaction Zn(s) + 2 HCl(aq) ZnCl 2 (aq) + H 2 (g) What mass of ZnCl 2 (MW = 136.3 g/mol) can be prepared from the reaction of 3.27 grams of zinc (MW = 65.38 g/mol) with 3.00 grams HCl (MW = 36.46)?

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