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Solving Stoichiometric Problems

Solving Stoichiometric Problems. Created by Suzanne Landrey Spring 2006. In this module, we’ll learn about:. Chemical equations Coefficients Balancing equations Stoichiometric ratios Using ratios to solve problems Molecule to molecule problems. reaction direction. reactants. 2.

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Solving Stoichiometric Problems

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  1. Solving Stoichiometric Problems Created by Suzanne Landrey Spring 2006

  2. In this module, we’ll learn about: • Chemical equations • Coefficients • Balancing equations • Stoichiometric ratios • Using ratios to solve problems • Molecule to molecule problems

  3. reaction direction reactants 2 H2(g) + F2(g) HF(g) stoichiometric coefficient Chemical Equations The same number of each type of atom appears on both sides of the equation, since mass is conserved during the reaction. We have 2 molecules of HF, so we must add a coefficient of 2 in front of the products in the chemical reaction. REMEMBER - Only alter the stoichiometric coefficients when balancing a chemical equation, not the subscripts in each formula.

  4. H2(g) + F2(g) 2HF(g) Chemical Equations To make sure your equation is balanced, count the number of each type of atom on each side of the equation to make sure they are equal. Two H atoms on the left side, two H atoms on the right. Two F atoms on the left side, two F atoms on the right. This equation is balanced!

  5. Chemical Equations Now it’s your turn to try a few practice problems on balancing equations. Just click on the correct answer and follow the directions. Good luck!

  6. Practice Problems – Balancing Equations 1. Which element is unbalanced in the following reaction? Na2SO4 + C  Na2S + CO Na S C O

  7. Practice Problems – Balancing Equations 1. Which element is unbalanced in the following reaction? Na2SO4 + C  Na2S + CO Na S C O Incorrect. The number of Na atoms is equal on each side of the equation. There is one Na atom on the left side of the equation and one on the right. Click here to try again.

  8. Practice Problems – Balancing Equations 1. Which element is unbalanced in the following reaction? Na2SO4 + C  Na2S + CO Na S C O Incorrect. The number of S atoms is equal on each side of the equation. There is one S atom on the left side of the equation and one on the right. Click here to try again.

  9. Practice Problems – Balancing Equations 1. Which element is unbalanced in the following reaction? Na2SO4 + C  Na2S + CO Na S C O Incorrect. The number of C atoms is equal on each side of the equation. There is one C atom on the left side of the equation and one on the right. Click here to try again.

  10. Practice Problems – Balancing Equations 1. Which element is unbalanced in the following reaction? Na2SO4 + C  Na2S + CO Na S C O Correct! There are 4 O atoms on the left side of the equation, but only one on the right side. Click here to try the next question.

  11. Practice Problems – Balancing Equations 2. Which coefficient belongs in front of the H2O in the following equation? Mg + ? H2O  Mg(OH)2 + H2 4 1 2 8

  12. Practice Problems – Balancing Equations 2. Which coefficient belongs in front of the H2O in the following equation? Mg + ? H2O  Mg(OH)2 + H2 4 1 2 8 Incorrect. Currently, there are a total of 4 H atoms on the right side of the equation, and 2 on the left. You do want there to be 4 H atoms on the left side of the equation, but if you put a coefficient of 4 in front of the H2O, there will be a total of 8 H atoms on the left side of the equation (4 x 2 = 8), meaning it will not be balanced. Click here to try again.

  13. Practice Problems – Balancing Equations 2. Which coefficient belongs in front of the H2O in the following equation? Mg + ? H2O  Mg(OH)2 + H2 4 1 2 8 Incorrect. Currently, there are a total of 4 hydrogen atoms on the right side of the equation, and 2 on the left. If you put a coefficient of 1 in from of the H2O, which is the same as not putting any coefficient in front of it, there will still be a total of 2 hydrogen atoms on the left side of the equation (1 x 2 = 8), meaning it will not be balanced. Click here to try again.

  14. Practice Problems – Balancing Equations 2. Which coefficient belongs in front of the H2O in the following equation? Mg + ? H2O  Mg(OH)2 + H2 4 1 2 8 Correct! There are a total of 4 H atoms on the right side of the equation, and 2 on the left. So by adding a coefficient of 2 in from of the H2O, there will be a total of 4 H atoms on left side of the equation as well, balancing it. Mg + 2 H2O  Mg(OH)2 + H2 Click here to try the next question.

  15. Practice Problems – Balancing Equations 2. Which coefficient belongs in front of the H2O in the following equation? Mg + ? H2O  Mg(OH)2 + H2 4 1 2 8 Incorrect. Currently, there are a total of 4 hydrogen atoms on the right side of the equation, and 2 on the left. If you put a coefficient of 8 in front of the H2O, there will be a total of 8 hydrogen atoms on the left side of the equation (4 x 2 = 8), meaning it will not be balanced. Click here to try again.

  16. Practice Problems – Balancing Equations 3. Balance the following equation. Na2SO4 + C  Na2S + CO Na2SO4 + 4C  Na2S + 4CO Na2SO4 + C  Na2S + 4CO Na2SO4 + C  Na2S + CO4

  17. Practice Problems – Balancing Equations 3. Balance the following equation. Na2SO4 + C  Na2S + CO Na2SO4 + 4C  Na2S + 4CO Na2SO4 + C  Na2S + 4CO Na2SO4 + C  Na2S + CO4 Correct! To begin with, there were:2 Na on the left and 2 Na on the right1 S on the left and 1 S on the right4 O on the left and 1 O on the right, and1 C on the left and 1 C on the right. So, all atoms are balances except the number of O atoms. Since there are 4 O on the left, if we put a 4 in front of the CO, we now have an equal number on both sides (4 on both sides). But then, the number of C atoms is different (1 on the left and 4 on the right from the coefficient of 4 we just added). Na2SO4 + C  Na2S + 4CO So we need to add a 4 in front of the C on the left. Now there are 4 C atoms on both sides and the equation is balanced. Na2SO4 + 4C  Na2S + 4CO Click here to move on to the next section – Stoichiometric Ratios.

  18. Practice Problems – Balancing Equations 3. Balance the following equation. Na2SO4 + C  Na2S + CO Na2SO4 + 4C  Na2S + 4CO Na2SO4 + C  Na2S + 4CO Na2SO4 + C  Na2S + CO4 Incorrect. If you add a 4 in front of the CO, it balances out the number of O atoms (4 on each side), but now the number of C atoms on each side is different (1 on the left and 4 on the right). Click here to try again.

  19. Practice Problems – Balancing Equations 3. Balance the following equation. Na2SO4 + C  Na2S + CO Na2SO4 + 4C  Na2S + 4CO Na2SO4 + C  Na2S + 4CO Na2SO4 + C  Na2S + CO4 Incorrect. When balancing equations, you cannot add or remove subscripts to atoms or molecules; you can only add coefficients in front of molecules. The subscripts are part of the chemical formula. They represent the structure of the molecule and the number of each element in the molecule, so changing them would mean changing the molecular structure. Coefficients can be added because they simply represent the number of molecules present. Click here to try again.

  20. Stoichiometric Ratios Now that we can balance equations, let’s look at the balanced HF equations we used earlier. If we added another H2 molecule and F2 molecule to our original equation, it would double the reaction.

  21. 1 molecule H2 2 molecules HF 2 molecules H2 4 molecules HF Stoichiometric Ratios So, we could say that for every H2 molecule, two HF molecules are produced: Or, for every two H2 molecules, four HF molecules are produced: These are called stoichiometric ratios and can be written like this: OR

  22. 1 molecule H2 2 molecules HF 2 molecules H2 4 molecules HF Stoichiometric Ratios One molecule of H2 to two molecules of HF. Or two molecules of H2 to four molecules of HF. These two ratios are the same, since it’s the same reaction. It’s just like a reduced fraction.

  23. 1 molecule H2 2 molecules HF 2 molecules HF 1 molecule H2 Stoichiometric Ratios Since these are ratios, we can invert them and they will still be correct. Or we can say We can use these ratios to solve chemistry problems. When solving stoichiometric problems, we must make sure we use the correct form of the ratio depending on the problem we are trying to solve. You may need to use the ratio 1 molecule H2for some problems, and 2 molecules HFfor others.2 molecules HF 1 molecule H2

  24. 1 molecule H2 2 molecules HF 2 molecules HF 1 molecule H2 Stoichiometric Ratios The ratio we need for a particular problem depends on what the known amounts in the problem are, and what the problem is asking you to find. For instance, if you have the number (or coefficient) of HF molecules in the equation and you need to find the number of H2 molecules that will be produced, you must use the ratio with H2 in the numerator. Example: During a reaction, 12 molecules of HF were produced. How many H2 molecules were present as reactants? But if you have the number of H2 molecules in the equation and you need to find the number of HF molecules that will be produced, you must use the ratio with HF in the numerator. Example: During a reaction, 7 molecules of H2 reacted to produce HF. How many molecules of HF were produced?

  25. 5H2(g) + 5F2(g) ? HF(g) 1 molecule H2 2 molecules HF 2 molecules HF 1 molecule H2 Stoichiometry – Molecule to Molecule Problems We can use these ratios to solve chemistry problems. For instance, we can figure out how many molecules of HF will be produced if there are 5 molecules of H2 and 5 molecules of F2 in the equation. We already know that for the HF equation, the ratio of H2 to HF is: And since we are looking for the number of HF molecules, we must use the ratio with HF in the numerator, and multiply it by 5. 5 molecules H2 = 10 molecules HF

  26. 10 molecules H2O 4 molecules C3H5N3O9 10 molecules H2O 4 molecules C3H5N3O9 4 molecules C3H5N3O9 10 molecules H2O If 20 molecules of nitroglycerin (C3H5N3O9) explode, how many molecules of water are produced? The reaction for this problem is: 4 C3H5N3O9 12 CO2 + 10 H2O + 6 N2 + O2 So, the ratios we want to use for this problem must involve nitroglycerin and water. We can see in the equation that there are 4 molecules of nitroglycerin to 10 molecules of water. Ratios: OR Since we need to find the number of water molecules produced, we must use the ratio with water in the numerator. Now, we multiply the ratio by 20 molecules of nitroglycerin. = 50 molecules H2O 20 C3H5N3O X

  27. 12 molecules CO2 4 molecules C3H5N3O9 6 molecules N2 4 molecules C3H5N3O9 1 molecule O2 4 molecules C3H5N3O9 In the same reaction, how many molecules of CO2, N2, and O2 are produced? Using the same reaction: 4 C3H5N3O9 12 CO2 + 10 H2O + 6 N2 + O2 But different ratios:

  28. Stoichiometric Problems Ok, it’s that time again to try a few more practice problems. Good luck!

  29. Practice Problems – Stoichiometric Problems 1. For the following reaction, what is the ratio of Fe molecules to Fe2O3 molecules? 4Fe + 3O2 2Fe2O31 molecules Fe 2 molecule Fe2O3 2 molecules Fe 1 molecule Fe2O3 4 molecules Fe 2 molecule Fe2O3 4 molecules Fe 3 molecule Fe2O3

  30. Practice Problems – Stoichiometric Ratios 1. For the following reaction, what is the ratio of Fe molecules to Fe2O3 molecules? 4Fe + 3O2 2Fe2O31 molecules Fe 2 molecule Fe2O3 2 molecules Fe 1 molecule Fe2O3 4 molecules Fe 2 molecule Fe2O3 4 molecules Fe 3 molecule Fe2O3 Incorrect. There are 4 molecules of Fe to 2 molecules of Fe2O3, so you must simplify that ratio : 4 = 2 2 1 Click here to try again.

  31. Practice Problems – Stoichiometric Ratios 1. For the following reaction, what is the ratio of Fe molecules to Fe2O3 molecules? 4Fe + 3O2 2Fe2O31 molecules Fe 2 molecule Fe2O3 2 molecules Fe 1 molecule Fe2O3 4 molecules Fe 2 molecule Fe2O3 4 molecules Fe 3 molecule Fe2O3 Correct! There are 4 molecules of Fe (on the left side of the equation) to 2 molecules of Fe2O3 (on the right side of the equation). So, the ratio would be: 4 molecules Fe2 molecules Fe2O3 then simplified: 4 = 2 so 4 molecules Fe 2 1 2 molecules Fe2O3 Click here to try the next question.

  32. Practice Problems – Stoichiometric Ratios 1. For the following reaction, what is the ratio of Fe molecules to Fe2O3 molecules? 4Fe + 3O2 2Fe2O31 molecules Fe 2 molecule Fe2O3 2 molecules Fe 1 molecule Fe2O3 4 molecules Fe 2 molecule Fe2O3 4 molecules Fe 3 molecule Fe2O3 Technically, this isn’t completely incorrect, but it’s not the best answer. You wouldn’t typically leave the ratio in this form. Like any other fraction, you should simplify it. Click here to try again (and this time, SIMPLIFY!).

  33. Practice Problems – Stoichiometric Ratios 1. For the following reaction, what is the ratio of Fe molecules to Fe2O3 molecules? 4Fe + 3O2 2Fe2O31 molecules Fe 2 molecule Fe2O3 2 molecules Fe 1 molecule Fe2O3 4 molecules Fe 2 molecule Fe2O3 4 molecules Fe 3 molecule Fe2O3 Incorrect. There are 4 molecules of Fe to 3 molecules of O2, but the question is asking for the ratio of Fe molecules to Fe2O3 molecules. Click here to try again.

  34. Practice Problems – Stoichiometric Ratios 2. For the following reaction, what is the ratio of NH3 molecules to H2 molecules? 3H2 + N2 2NH3 2 molecules NH3 1 molecules H2 3 molecules NH3 2 molecules H2 2 molecules NH3 3 molecules H2 3 molecules NH3 1 molecules H2

  35. Practice Problems – Stoichiometric Ratios 2. For the following reaction, what is the ratio of NH3 molecules to H2 molecules? 3H2 + N2 2NH3 2 molecules NH3 1 molecules H2 3 molecules NH3 2 molecules H2 2 molecules NH3 3 molecules H2 3 molecules NH3 1 molecules H2 Incorrect. There are 2 molecules of NH3 (on the right side of the equation) to 1 molecule of N2 (on the left side of the equation), but the question is asking for the ratio of NH3 molecules to H2 molecules. Click here to try again.

  36. Practice Problems – Stoichiometric Ratios 2. For the following reaction, what is the ratio of NH3 molecules to H2 molecules? 3H2 + N2 2NH3 2 molecules NH3 1 molecules H2 3 molecules NH3 2 molecules H2 2 molecules NH3 3 molecules H2 3 molecules NH3 1 molecules H2 Incorrect. There are 2 molecules of NH3 (on the right side of the equation) to 3 molecules of H2 (on the left side of the equation). Click here to try again.

  37. Practice Problems – Stoichiometric Ratios 2. For the following reaction, what is the ratio of NH3 molecules to H2 molecules? 3H2 + N2 2NH3 2 molecules NH3 1 molecules H2 3 molecules NH3 2 molecules H2 2 molecules NH3 3 molecules H2 3 molecules NH3 1 molecules H2 Correct! There are 2 molecules of NH3 (on the right side of the equation) to 3 molecules of H2 (on the left side of the equation). Click here to try the next question.

  38. Practice Problems – Stoichiometric Ratios 2. For the following reaction, what is the ratio of NH3 molecules to H2 molecules? 3H2 + N2 2NH3 2 molecules NH3 1 molecules H2 3 molecules NH3 2 molecules H2 2 molecules NH3 3 molecules H2 3 molecules NH3 1 molecules H2 Incorrect. There are 3 molecules of H2 (on the right side of the equation) to 1 molecule of N2, but the question is asking for the ratio of NH3 molecules to H2 molecules. Click here to try again.

  39. Practice Problems – Stoichiometric Ratios 3. In the following reaction, how many molecules of NH3 will be produced if 9 molecules of H2 are reacting? 3H2 + N2 2NH3 8 9 3 6

  40. Practice Problems – Stoichiometric Ratios 3. In the following reaction, how many molecules of NH3 will be produced if 9 molecules of H2 are reacting? 3H2 + N2 2NH3 8 9 3 6 Incorrect. Instead of adding 6 to both the NH3 and H2 molecules in this equation, you must multiply both by 3 to get the correct number. Click here to try again.

  41. Practice Problems – Stoichiometric Ratios 3. In the following reaction, how many molecules of NH3 will be produced if 9 molecules of H2 are reacting? 3H2 + N2 2NH3 8 9 3 6 Incorrect. There are 2 molecules of NH2 (on the right side of the equation) to 3 molecules of H2 (on the left side of the equation), so if the number of H2 molecules changes to 9, the number of NH2 molecules can’t also be 9. Click here to try again.

  42. Practice Problems – Stoichiometric Ratios 3. In the following reaction, how many molecules of NH3 will be produced if 9 molecules of H2 are reacting? 3H2 + N2 2NH3 8 9 3 6 Incorrect. There are 2 molecules of NH2 (on the right side of the equation) to 3 molecules of H2 (on the left side of the equation), so you must use the following ratio for this problem: 2 molecules NH33 molecules H2 Since you are trying to find the number of NH3 produced when 9 molecules of H2 is used (instead of 3 molecules of H2), you must multiply the ratio above by 3: Click here to try again.

  43. Practice Problems – Stoichiometric Ratios 3. In the following reaction, how many molecules of NH3 will be produced if 9 molecules of H2 are reacting? 3H2 + N2 2NH3 8 9 3 6 Correct! There are 2 molecules of NH3 (on the right side of the equation) to 3 molecules of H2 (on the left side of the equation), so you use the following ratio for this problem: 2 molecules NH33 molecules H2 Since you are trying to find the number of NH3 produced when 9 molecules of H2 is used (instead of 3 molecules of H2), you must multiply the ratio above by 3: So, now there are 6 molecules of NH3 to 9 molecules of H2. Click here to review and take the end of lesson quiz.

  44. Review So now that you have completed this module, you know all about: • Chemical equations and coefficients • Stoichiometric ratios • How to solve stoichiometric problems using ratios.And you’re ready to take the end of lesson quiz.Click here to take the quiz.

  45. End of Lesson Quiz 1. During a reaction, 24 molecules of FeO2 is produced. 4Fe + 3O2 2Fe2O3 How many molecules of O2 were present in the reaction? 2. During a reaction, 75 molecules of NO reacts to produce NO2. 2NO + O2 2NO2 How many molecules of O2 were used in the reaction? 3. In the following reaction, how many molecules of K2O will be produced if 47 molecules of O2 are reacting? 2K2S + O2 2K2O + 2S

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