1 / 8

Bell Ringer

Calculate the grams of carbon dioxide produced when 25.0 liters of butane (C4H10) is burned. Use volume conversions and stoichiometry.

jhickson
Download Presentation

Bell Ringer

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 13 C C 2 H H O O Bell Ringer 25.0 liters of butane (C4H10) are burned. How many grams of carbon dioxide are obtained? 2 C4H10 + 13 ? O2 4 8 CO2 + 10 5 ? H2O x 2 x 2 4 8 1 4 8 10 20 2 10 20 2 26 3 13 26

  2. Bell Ringer 25.0 liters of butane (C4H10) are burned. How many grams of carbon dioxide are obtained? 2 C4H10 + 13 ? O2 4 8 CO2 + 10 5 H2O 1 mol C4H10 8 mol CO2 44.01 g CO2 25.0 L C4H10 x x x = 22.4 L C4H10 2 mol C4H10 1 mol CO2 196 g CO2

  3. Volume Conversions Quiz 2 C4H10 + 13 ? O2 8 4 CO2 + 10 5 H2O 3 points 1 mol C4H10 8 mol CO2 44.01 g CO2 25.0 L C4H10 x x x = 22.4 L C4H10 2 mol C4H10 1 mol CO2 2 points 3 points 2 points 2 Questions, 10 points each - watch significant figures - make sure labels cancel… label EVERYTHING!

  4. Volume Conversions Quiz Good Luck!!

  5. Homework Answers • O2 is the limiting reactant. • 0.90 g H2 are left over. • Fe is the limiting reactant. • 0.19 g S are left over. • There was a 78.4 % yield of NO gas.

  6. 2 2 Homework #21 Identify the limiting reactant when 1.22 g of O2 reacts with 1.05 g of H2 to produce water. O2 + H2 H2O 1 mol O2 2 mol H2O 18.02 g H2O 1.37 g H2O 1.22 g O2 x x x = 32.00 g O2 1 mol O2 1 mol H2O LIMITING REACTANT THEORETICAL YIELD 1 mol H2 2 mol H2O 18.02 g H2O 9.36 g H2O 1.05 g H2 x x x = EXCESS REACTANT 2.02 g H2 2 mol H2 1 mol H2O

  7. 2 2 Homework #22 Determine the amount of excess reactant for question 21. Information from previous problem: Limiting Reactant: O2 (1.22 g) Excess Reactant: H2 (1.05 g) AMOUNT OF EXCESS REACTANT USED O2 + H2 H2O 1 mol O2 2 mol H2 2.02 g H2 0.154 g H2 1.22 g O2 x x x = 32.00 g O2 1 mol O2 1 mol H2 1.05 g H2 0.154 g H2 0.90 g H2

  8. % YIELD = ACTUAL YIELD x 100% THEORETICAL YIELD Homework #25 Determine the percent yield for the reaction between 15.8 g of NH3 and excess oxygen that has an actual yield of 21.8 g of NO gas and water. 4 2 NH3 + 5/2 5 O2 2 4 NO (g) + 6 3 H2O 1 mol NH3 4 mol NO 30.01 g NO 15.8 g NH3 x x x = 17.04 g NH3 4 mol NH3 1 mol NO 27.8 g NO THEORETICAL YIELD 21.8 g NO (actual) = x 100% = 78.4 % 27.8 g NO (theoretical)

More Related