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Chapter 12

Stoichiometry. Chapter 12. Stoichiometry. Goal Calculate quantities of reactants and products needed in chemical rxns using balanced chemical equations. Using an Equation as a Recipe . Ingredients: F rame, S eat, W heels, H andlebar, P edals Balanced equation:

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Chapter 12

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  1. Stoichiometry Chapter 12

  2. Stoichiometry Goal • Calculate quantities of reactants and products needed in chemical rxns using balanced chemical equations.

  3. Using an Equation as a Recipe • Ingredients: Frame, Seat, Wheels, Handlebar, Pedals • Balanced equation: 1F + 1S + 2W + 1H + 2P 1 FSW2HP2

  4. How many pedals are needed to make 128 bicycles? 128 FSW2HP2 2P = 256 pedals 1 FSW2HP2 1F + 1S + 2W + 1H + 2P 1 FSW2HP2

  5. Examining a Balanced Chemical Equation 2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(g) • 2 moles of H2S reacts with 3 moles of O2 to produce 2 moles of SO2 and 2 moles of H2O. • 4 moles of H2S reacts with ___moles of O2 • ____ moles of O2 are needed to produce 7 moles of H2O • How to solve these is on the next slide. 6 10.5

  6. 4 molH2S 7molH2O 3 mole O2 3 mole O2 = 6 mol O2 = 10.5 mol O2 2 mol H2S 2 mol H2O Mole-Mole Ratio Problems The mol-mol ratio comes from the coefficients in the balanced equation. In the balance equation, O2 has a coefficient of 3, H2O has a coefficient of 2, and H2S has a coefficient of 2.

  7. Mass-MassCalculation • Write the balanced chemical equation • Label above the equation what you are given and what you are trying to find. • Convert grams of your given to moles using molar mass • Multiply by the mole-mole ratio from the balanced chemical equation • Convert moles of unknown to grams

  8. Bringing It All Together • How many grams of ammonia ( ) should be produced when 6.00g nitrogen react with excess hydrogen? • Solve: Write the balanced equation with the information labeled above it. 6.00 g excess ? g N2 + 3H2 2NH3 NH3

  9. 6.00gN2 1 mole N2 = 0.214 mol N2 28 g N2 Bringing It All Together • Convert given to moles: • Multiply by the mole ratio: 0.214 mol N2 2 mol NH3 = 0.428 mol NH3 1 mol N2

  10. Wrapping It Up Convert moles of unknown to grams: 0.428 mol NH3 17g NH3 = 7.28g NH3 1 mol NH3 Summary: When 6.00g of nitrogen(N2) reacts with an excess of Hydrogen (H2), 7.28g of ammonia (NH3) should be produced.

  11. Now You Try! What mass of aluminum is required to completely react with 7.80g oxygen in a synthesis reaction?

  12. Write the balanced equation 4Al + 3O2 2Al2O3

  13. Label your given and unknown information Since you were not given info about this compound, don’t worry about it and don’t use it. 4Al + 3O2 2Al2O3 7.80 g

  14. NEXT?... • Convert given to moles: • Multiply by the mole ratio: 7.80g O2 1 mol O2 = 0.244mol O2 32g O2 0.244mol O2 4 mol Al = 0.325 mol Al 3 mol O2

  15. AND FINALLY!! Convert moles of unknown to grams: 0.325 mol Al 27g Al = 8.78g Al 1 mol Al 8.78g of aluminum (Al) should be required to react completely with7.80g of oxygen (O2) in the synthesis of aluminum oxide (Al2O3).

  16. Mass-Volume Calculation • Write the balanced chemical equation • Label above the equation what you are given and what you are trying to find. • Convert grams of your given to moles using molar mass • Multiply by the mole-mole ratio from the balanced chemical equation • Convert moles of unknown to liters (a.k.a dm3) using molar volume (22.4L/ 1mol)

  17. Bringing It All Together • How many liters of oxygen are necessary for the combustion of 134g of magnesium, assuming the reactions occurs at STP? • Solve: Write the balanced equation with the information labeled above it. 134 g ? L 2Mg+ O2 2MgO

  18. Bringing It All Together • Convert given to moles: • Multiply by the mole ratio: 134gMg 1 mole Mg = 5.514 mol Mg 24.3 g Mg 5.514 mol Mg 1 mol O2 = 2.757 mol O2 2 mol Mg

  19. Wrapping It Up Convert moles of unknown to liters: 2.757 molO2 22.4 LO2 = 61.8 L O2 1 mol O2 61.8 L of O2 are required to completely combust 134g Mg in the synthesis of MgO.

  20. Volume-Mass Calculation • Write the balanced chemical equation • Label above the equation what you are given and what you are trying to find. • Convert liters of your given to moles using molar volume • Multiply by the mole-mole ratio from the balanced chemical equation • Convert moles of unknown to grams using molar mass

  21. Bringing It All Together • Find the mass of sulfur (S8) required to react with oxygen to produce 2.47L of sulfur dioxide gas at STP. • Solve: Write the balanced equation with the information labeled above it. ? g 2.47L S8 + 8 O2 8SO2

  22. Bringing It All Together • Convert given to moles: • Multiply by the mole ratio: 2.47LSO2 1 mole SO2 = 0.1103 mol SO2 22.4 L SO2 0.1103 mol SO2 1 mol S8 = 0.01378 mol S8 8 mol SO2

  23. Wrapping It Up Convert moles of unknown to grams: 0.01378 mol S8 256g S8 = 3.53 g S8 1 mol S8 3.53g of sulfur (S8) are required to produce 2.47 L of sulfur dioxide (SO2).

  24. Volume-Volume Calculations • Write the balanced chemical equation • Label above the equation what you are given and what you are trying to find. • Convert liters of your given to moles using molar volume (22.4L/1mol) • Multiply by the mole-mole ratio from the balanced chemical equation • Convert moles of unknown to liters using molar volume (22.4L/1 mol)

  25. Bringing It All Together • What volume of H2S gas is needed to produce 14.2 L of water at STP? • Solve: Write the balanced equation with the information labeled above it. ?L 14.2L 2H2S+ 3O2 2SO2 + 2 H2O

  26. Bringing It All Together • Convert given to moles: • Multiply by the mole ratio: 14.2 LH2O 1 mole H2O = 0.6339 mol H2O 22.4 L H2O 0.6339 mol H2O 2 mol H2S = 0.6339 mol H2S 2mol H2O

  27. Wrapping It Up Convert moles of unknown to liters: 0.6993 mol H2S 22.4 LH2S = 14.2 L H2S 1 mol H2S 14.2 L of H2S are required to produce 14.2 L H2O.

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