# HW5 - momin(rrm497 – Homework 5 – Sutcliffe –(53120 1...

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Unformatted text preview: momin (rrm497) – Homework 5 – Sutcliffe – (53120) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points For the reaction 2 H 2 + O 2 → 2 H 2 O , what is the maximum amount of H 2 O which could be formed from 16 . 37 g of H 2 and . 175 mol of O 2 ? Correct answer: 6 . 30535 g. Explanation: m H 2 = 16 . 37 g n O 2 = 0 . 175 mol This is a limiting reactant problem because the amounts of more than one reactant are given. Either H 2 or O 2 will limit the amount of H 2 O that can form. Assume that H 2 is the limiting reactant. The amount of H 2 O that can be produced is ? g H 2 O = 16 . 37 g H 2 × 1 mol H 2 2 . 01588 g H 2 × 2 mol H 2 O 2 mol H 2 × 18 . 0153 g H 2 O 1 mol H 2 O = 146 . 293 g H 2 O . Assume that O 2 is the limiting reactant. The amount of H 2 O that can be produced is ? g H 2 O = 0 . 175 mol O 2 × 2 mol H 2 O 1 mol O 2 × 18 . 0153 g H 2 O 1 mol H 2 O = 6 . 30535 g H 2 O . Since a smaller amount of H 2 O can be pro- duced with the given amount of O 2 , O 2 is the limiting reagent, and a maximum of 6 . 30535 g of H 2 O can be produced. 002 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( ℓ ) is carried out to completion, 1. all of the O 2 is consumed and 4.0 mol of NO(g) is produecd. 2. 1.5 mol of H 2 O is produced. 3. all of the O 2 is consumed. correct 4. all of the NH 3 is consumed. 5. all of the NH 3 is consumed and 1.5 mol of H 2 O is produced. 6. all of the O 2 is consumed and 1.5 mol of NO(g) is produced. 7. all of the NH 3 is consumed and 4.0 mol of NO(g) is produced. 8. all of the O 2 is consumed and 1.5 mol of H 2 O is produced. 9. 4.0 mol of NO(g) is produced and 1.5 mol of H 2 O is produced. 10. None of the other answers is correct. 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 = . 8 mol NH 3 1 mol O 2 From this ratio we see that each mole of O 2 that reacts requires exactly 0.8 mol NH 3 . Next momin (rrm497) – Homework 5 – Sutcliffe – (53120) 2 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....
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## This note was uploaded on 12/11/2009 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas.

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HW5 - momin(rrm497 – Homework 5 – Sutcliffe –(53120 1...

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