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Stoichiometry Practice Sheet

Stoichiometry Practice Sheet. Init 1/10/2012 by Daniel R. Barnes.

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Stoichiometry Practice Sheet

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  1. Stoichiometry Practice Sheet Init 1/10/2012 by Daniel R. Barnes WARNING: This presentation may contain graphical images and other content that was taken from the world wide web without permission of the owner(s). Please do not copy or distribute this presentation. Its very existence may be illegal.

  2. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 1. Draw a cartoon of the reaction below, represeting each atom as a circle with its element’s symbol inside. O O C O C H H H O C O O C C C H H O O H H H H H O O O H H H O O H O O O O O O H H O

  3. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 1 2. How many oxygen molecules are consumed for every propane molecule burned? used up A: five Would looking at the cartoon help? H H H O O O C C C C C C H H O O O O H H H O O O H H H H O O O O O O H H O H H O O O

  4. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 1 2. How many oxygen molecules are consumed for every propane molecule burned? A: five Would looking at the cartoon help? One molecule of propane H H H O O O C C C C C C H H O O O O H H H 4 O O O 1 3 H H H H O O O O O O 5 2 H H O H H O O O Five molecules of oxygen

  5. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 1 3. How many carbon dioxide molecules are produced for every propane molecule burned? A: three Would looking at the cartoon help? H H H O O O C C C C C C H H O O O O H H H O O O H H H H O O O O O O H H O H H O O O

  6. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 1 3. How many carbon dioxide molecules are produced for every propane molecule burned? Three molecules of carbon dioxide A: three One molecule of propane H H H O O O C C C C C C H H O O O O 1 2 3 H H H O O O H H H H O O O O O O H H O H H O O O

  7. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 4. How many water molecules are created for every oxygen molecule consumed? 4 water molecules = 0.8 water molecules/oxygen molecule 5 oxygen molecules

  8. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 5. If two million propane molecules are burned, how many water molecules are produced? 4 water molecules 2 million propane molecules x propane molecule 1 = 8 million water molecules

  9. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 6. How many oxygen molecules does it take to produce nine billion carbon dioxide molecules? 5 oxygen molecules 9 billion carbon dioxide molecules x 3 carbon dioxide molecules 1 = 15 billion oxygen molecules

  10. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 7. How many propane molecules does it take to produce 484 dozen water molecules? 1 propane molecules 484 dozen water molecules x 4 water molecules 1 = 121 dozen propane molecules

  11. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 8. How much carbon dioxide would be produced by burning 10 kazillion oxygen molecules? 3 carbon dioxide molecules 10 kazillion oxygen molecules x 5 oxygen molecules 1 = 6 kazillion carbon dioxide molecules

  12. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 9. For every mole of carbon dioxide produced, how many moles of water are produced? 4 mol water 1 mol carbon dioxide x 3 mol carbon dioxide 1 = 1.33 mol water You could also say, “1.33 mol water/mol carbon dioxide.”

  13. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 10. What is the oxygen-to-carbon dioxide ratio in this reaction? 5 oxygen molecules 5 to 3 = 5:3 = 3 carbon dioxide molecules 5 mol oxygen = 3 mol carbon dioxide

  14. Propane combusts in the presence of oxygen (burns) according to the following balanced equation: C3H8 + 5O2 3CO2 + 4H2O 11. In general, how do you figure out the mole ratio of any two substances in a chemical reaction? The coefficients COEFFICIENTS MOLE RATIO

  15. And now, for the BACK SIDE of the worksheet (Sorry. The rest of the front side is under construction for now.)

  16. 12. What mass of magnesium oxide should be produced when 12.15 grams of magnesium metal burns? Magnesium burns according to the following equation: 2Mg + O2 2MgO g 20 12.15 g 2Mg + O2 2MgO 12.15 g Mg 1 mol Mg 2 mol MgO 40 g MgO 1 24 g Mg 2 mol Mg 1 mol MgO Mg: MgO: Mg: 1 x 24 = 24 Mg: 1 x 24 = 24 O: 1 x 16 = 16 24 g/mol 40 g/mol 960 12.15 x 2 x 40 g MgO g MgO g MgO = 20 = 48 24 x 2 Just show me the answer!

  17. 12. What mass of magnesium oxide should be produced when 12.15 grams of magnesium metal burns? Magnesium burns according to the following equation: 2Mg + O2 2MgO g 20 12.15 g 2Mg + O2 2MgO 12.15 g Mg 1 mol Mg 2 mol MgO 40 g MgO 1 24 g Mg 2 mol Mg 1 mol MgO Mg: MgO: Mg: 1 x 24 = 24 Mg: 1 x 24 = 24 O: 1 x 16 = 16 24 g/mol 40 g/mol 960 12.15 x 2 x 40 g MgO g MgO g MgO = 20 = 48 24 x 2

  18. 13. How many grams of hydrogen gas can 128 grams of oxygen burn? Hydrogen gas burns according to the following balanced equation: 2H2 + O2 2H2O. g 16 128 g 2H2 + O2 2H2O 128 g O2 1 mol O2 2 mol H2 2 g H2 1 32 g O2 1 mol O2 1 mol H2 O2: H2: O: 2 x 16 = 32 H: 2 x 1 = 2 2 g/mol 32 g/mol 512 128 x 2 x 2 g H2 g H2 g H2 = 16 = 32 32 Just show me the answer!

  19. 13. How many grams of hydrogen gas can 128 grams of oxygen burn? Hydrogen gas burns according to the following balanced equation: 2H2 + O2 2H2O. g 16 128 g 2H2 + O2 2H2O 128 g O2 1 mol O2 2 mol H2 2 g H2 1 32 g O2 1 mol O2 1 mol H2 O2: H2: O: 2 x 16 = 32 H: 2 x 1 = 2 2 g/mol 32 g/mol 512 128 x 2 x 2 g H2 g H2 g H2 = 16 = 32 32

  20. 14. What mass of copper (II) oxide should be produced when 620 g copper (II) carbonate decomposes according to the following balanced equation: CuCO3 CuO + CO2 ? g 400 620 g CuCO3  CuO + CO2 620 g CuCO3 mol CuCO3 1 mol CuO 80 g CuO 1 1 124 g CuCO3 1 mol CuCO3 1 mol CuO CuO: CuCO3: Cu: 1 x 64 = 64 Cu: 1 x 64 = 64 C: 1 x 12 = 12 O: 1 x 16 = 16 O: 3 x 16 = 48 80 g/mol 124 g/mol 49,600 620 X 80 g CuO g CuO g CuO = 400 = Just show me the answer! 124 124

  21. 14. What mass of copper (II) oxide should be produced when 620 g copper (II) carbonate decomposes according to the following balanced equation: CuCO3 CuO + CO2 ? g 400 620 g CuCO3  CuO + CO2 620 g CuCO3 1 mol CuCO3 1 mol CuO 80 g CuO 1 124 g CuCO3 1 mol CuCO3 1 mol CuO CuO: CuCO3: Cu: 1 x 64 = 64 Cu: 1 x 64 = 64 C: 1 x 12 = 12 O: 1 x 16 = 16 O: 3 x 16 = 48 80 g/mol 124 g/mol 49,600 620 X 80 g CuO g CuO g CuO = 400 = 124 124

  22. 15. How many grams of NI3 must explode in order to produce 7620 grams of iodine vapors? Nitrogen triiodide decomposes according to the following balanced equation: 2NI3 N2 + 3I2 g 7900 7620 g 2NI3 N2 + 3I2 7620 g I2 mol I2 2 mol NI3 395 g NI3 1 1 254 g I2 3 mol I2 1 mol NI3 NI3: I2: I: 2 x 127 = 254 N: 1 x 14 = 14 I: 3 x 127 = 381 254 g/mol 395 g/mol 7620 x 2 x 395 6,019,800 g NI3 g NI3 g NI3 = 7900 = Just show me the answer! 762 254 x 3

  23. 15. How many grams of NI3 must explode in order to produce 7620 grams of iodine vapors? Nitrogen triiodide decomposes according to the following balanced equation: 2NI3 N2 + 3I2 g 7900 7620 g 2NI3 N2 + 3I2 7620 g I2 mol I2 2 mol NI3 395 g NI3 1 1 254 g I2 3 mol I2 1 mol NI3 NI3: I2: I: 2 x 127 = 254 N: 1 x 14 = 14 I: 3 x 127 = 381 254 g/mol 395 g/mol 7620 x 2 x 395 6,019,800 g NI3 g NI3 g NI3 = 7900 = 762 254 x 3

  24. HONORS PROBLEM (From the Q2 BMK, given 1/28-29/2014, by Action Learning Systems) • Iron reacts with oxygen to form rust, • 4Fe (s) + 3O2 (g)  2Fe2O3 (s). • When 56 g of iron (Fe) reacts with 96 g of oxygen (O2), how much rust (Fe2O3) will form? This is a limiting reactant stoichiometry problem. It’s stoichiometry because you are given grams of one chemical and asked to determine grams of some other chemical. It’s a limiting reactant problem because you are actually given grams for TWO chemicals, iron and oxygen. The way to solve this is to do two stoichiometry calculations. First, calculate how much rust can be made from 56 grams of iron. Then, calculate how much rust can be made from 96 grams of oxygen. The answer is the lesser of the two results, since oxygen and iron must cooperate to make rust, and a chain is only as strong as its weakest link.

  25. 96 g O2 56 g Fe mol O2 mol Fe 2 2 mol Fe2O3 mol Fe2O3 160 160 g Fe2O3 g Fe2O3 1 1 1 1 56 32 g O2 g Fe 3 4 mol O2 mol Fe 1 1 mol Fe2O3 mol Fe2O3 HONORS PROBLEM (From the Q2 BMK, given 1/28-29/2014, by Action Learning Systems) • Iron reacts with oxygen to form rust, • 4Fe (s) + 3O2 (g)  2Fe2O3 (s). • When 56 g of iron (Fe) reacts with 96 g of oxygen (O2), how much rust (Fe2O3) will form? Limiting reactant = 80 g Fe2O3 Excess reactant = 320 g Fe2O3

  26. If you think there’s something wrong or missing, please E-mail me! barnesd@centinela.k12.ca.us

  27. TM HYPERINDEX Click a button. Go to a place. #1 #5 #9 #12 Q2 BMK #38 bonus question #2 #6 #10 #13 #3 #7 #11 #14 #4 #8 #15

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