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Greek for “measuring elements” The calculations of quantities in chemical reactions based on a balanced equation.

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Mole to Mole conversions How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 2 Al 2 O 3 Al + 3 O 2 3.34 mol Al 2 O 3 2 mol Al 2 O 3 3 mole O 2 = 5.01 moles O 2 Comparing one coefficient to another is called using the molar ratio. The molar ratio in the balanced chemical equation remains constant.

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Your Turn 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed?

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We can’t measure moles!! We can convert grams to moles. –periodic table Then do the math with the moles. –balanced equation

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For example... If 10.1 g of iron are added to a solution of copper (II) sulfate, how many moles of copper would form? Fe (s) + CuSO 4(aq) Fe 2 (SO 4 ) 3(aq) + Cu (s) 233

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2Fe (s) + 3CuSO 4(aq) Fe 2 (SO 4 ) 3(aq) + 3Cu (s) 10.1 g Fe 55.85 g Fe 1 mol Fe = 0.18mol Fe 2 mol Fe 3 mol Cu =0.271 mol Cu

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We can’t measure moles!! We can convert grams to moles. –periodic table Then do the math with the moles. –balanced equation Then turn the moles back to grams. –periodic table

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For example... If 10.1 g of iron are added to a solution of copper (II) sulfate, how much solid copper would form? 2 Fe (s) + 3 CuSO 4(aq) Fe 2 (SO 4 ) 3(aq) + 3 Cu (s) 10.1 g Fe 55.85 g Fe 1 mol Fe = 0.181 mol Fe

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2Fe (s) + 3CuSO 4(aq) Fe 2 (SO 4 ) 3(aq) + 3Cu (s) 0.181 mol Fe 2 mol Fe 3 mol Cu = 0.272 mol Cu 1 mol Cu 63.55 g Cu =17.3 g Cu

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Could have done it 10.1 g Fe 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu =17.3 g Cu

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More Examples To make silicon for computer chips they use this reaction SiCl 4(l) + 2 Mg (s) 2 MgCl 2(aq) + Si (s) How many grams of Mg are needed to make 9.3 g of Si? How many grams of SiCl 4 are needed to make 9.3 g of Si? How many grams of MgCl 2 are produced along with 9.3 g of silicon?

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For Example The U. S. Space Shuttle boosters use this reaction 3Al (s) + 3NH 4 ClO 4(aq) Al 2 O 3(s) + AlCl 3(s) + 3NO (g) + 6H 2 O (l) How much Al must be used to react with 652 g of NH 4 ClO 4 ? How much water is produced? How much AlCl 3 ?

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2 Na (s) + Cl 2(g) → 2 NaCl (aq)

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Limiting Reagent If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make. The limiting reagent is the reactant you run out of first. The excess reagent is the one you have left over. The limiting reagent determines how much product you can make

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How do you find out? Do two stoichiometry problems. The one that makes the least product is the limiting reagent. For example Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

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If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? 2 Cu (s) + S (s) Cu 2 S (s) 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 159.16 g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 mol Cu 2 S 159.16 g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is Limiting Reagent

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Your turn If 10.1 g of magnesium and 2.87 g of HCl gas are reacted, how many grams of gas will be produced? How much excess reagent remains?

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Your Turn II If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper will be produced? How much excess reagent will remain?

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Yield The amount of product made in a chemical reaction. There are three types Actual yield- what you get in the lab when the chemicals are mixed Theoretical yield- what the balanced equation tells you you should make. Percent yieldPercent yield = Actual x 100 % Theoretical

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Example 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2 Al (s) + 3 CuSO 4(aq) Al 2 (SO 4 ) 3(aq) + 3 Cu (s) What is the actual yield? What is the theortical yield? What is the percent yield?

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Details Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %.

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Stoichiometry Chapter 9 Stoichiometry Greek for “measuring elements” The calculations of quantities in chemical reactions based on a balanced equation.

Stoichiometry Chapter 9 Stoichiometry Greek for “measuring elements” The calculations of quantities in chemical reactions based on a balanced equation.

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