d 25777167 x 1kg NH3 1000g NH3 e 25777kg NH3 Reference

D 25777167 x 1kg nh3 1000g nh3 e 25777kg nh3 reference

This preview shows page 4 - 7 out of 7 pages.

756772x2molNH3/1molNH3 d. 25777167 x 1kg NH3 / 1000g NH3 e. =25777kg NH3 (Reference: Chang 3.77) 7. Nitric acid (NO) reacts with oxygen gas to form nitrogen dioxide (NO 2 ), a dark brown gas: 2NO( g ) + O 2 ( g ) → 2NO 2 ( g ) In one experiment 0.886 mole of NO is mixed with 0.503 mol of O 2 . a. Calculate which of the two reactants is the limiting reagent. (5 points) b. Calculate also the number of moles of NO 2 produced. (5 points) c. What reactant is left over and how much of it is left over? (5 points) i. 0.886 mol NO x 2 mol 2 NO / 2 mol NO = 0.866 mol 2 NO ii. 0.503 mol 2 O x 2 mol 2 NO / MOL 2 O = 1.01 MOL 2 NO iii. NO is the limiting reagent. It limits the amount of product produced. The amount of product produced is 0.886 mol 2 NO iv. Reactant oxygen gas (O2) is left. The amount is (0.503 – 0.886 / 2) = 0.06 mole oxygen are left. (Reference: Chang 3.83) 4 Copyright © 2017 by Thomas Edison State University . All rights reserved.
Image of page 4
8. Propane (C 3 H 8 ) is a component of natural gas and is used in domestic cooking and heating. It burns according to the following reaction: C 3 H 8 + O 2 CO 2 + H 2 O a. Balance the equation representing the combustion of propane in air. (5 points) b. How many grams of carbon dioxide can be produced by burning 20.0 pounds of propane, the typical size of a BBG grill propane tank? Assume that oxygen is the excess reagent in this reaction. (5 points) i. 3 8 2 2 2 C H 5O 3CO 4H O ii. 1/3 = Moles of 3 8 C H / Moles of 2 CO iii. The Volume of One mole is 22.4 liters. iv. Number of Moles = Mass / Mass of 1 mole v. To determine the mass of 1 mole of 3 8 C H and 2 CO vi. For 3 8 C H mass = 3x12 +8 = 44 grams vii. For 2 CO Mass = 12 +2 x 16 = 44 grams 1. Convert 20 pounds to grams 2. Mass = 20lb x 454 g/lb = 9080 grams 3. Mass of 1 mole = 3x12+8=44 grams 4. Number of moles = 9080 / 44 5. 1/3 = 9080 / 44 / Moles of 2 CO 6. Moles of 2 CO = 3x9080 / 44 7. Approximately 619 moles 8. Mass = 44 x3 x 9080 / 44 =27,240 grams = 4g 2.72x10 of 2 CO 5 Copyright © 2017 by Thomas Edison State University . All rights reserved.
Image of page 5
9.
Image of page 6
Image of page 7

You've reached the end of your free preview.

Want to read all 7 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture