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Published byDorthy Montgomery Modified over 9 years ago
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Continuing Stoichiometry…
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The idea. In every chemical reaction, there is one reactant that will be run out (called the limiting reactant). This will limit the amount of product that can form. The reaction will stop at that point. There are then also excess reactants.
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Sample Problem S 8 + 4 Cl 2 4 S 2 Cl 2 If you have 200.0 g S 8 and 100.0 g Cl 2, what is the limiting reactant?
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Step 1 First, find the moles of each reactant. Amount of each reactant will be given Usually it will be in grams. This is a gram to mole conversion.
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The Answers 200.0 g S 8 1 mole S 8 x =.78 mol S 8 1 256 g S 8 100.0 g Cl 2 1 mole Cl 2 x = 1.41 mol Cl 2 1 71 g Cl 2
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If moles are Given… If moles are given, problem starts here…
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Step 2: Compare Mole Ratios A) What is the mole to mole ratio from the balanced chemical equation? S 8 + 4 Cl 2 4 S 2 Cl 2 Moles S 8 = 1 Moles Cl 2 = 4 Need 4 Cl 2 for each S 8 !
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Step 2 cont’d B) What is the mole to mole ratio calculated from the starting conditions? Moles S 8 =.78 mol S 8 Moles Cl 2 = 1.41 mol Cl 2 1.41 mol Cl 2 = 1.81 mol Cl 2 for 1 mol S 8.78 mol S 8
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Is this enough? NO! Therefore, Cl 2 is the limiting reactant Now practice some LR calculations!
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Calculating the Amount of Product Formed Complete a mole to gram calculation using the Limiting Reactant In our example: LR: Cl 2 = 1.41 moles Try it!
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The Answer S 8 + 4 Cl 2 4 S 2 Cl 2 1.41 mol Cl 2 4 mol S 2 Cl 2 135 g S 2 Cl 2 x x 1 4 mol Cl 2 1 mol S 2 Cl 2 = 190.35 g S 2 Cl 2
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Analyzing the Excess Reactant You care about two things: How much of the excess reactant reacted How much of the excess reactant remains
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Excess that Reacted Complete a mole to gram calculation using the limited and excess reactants In our example: Limiting Reactant: Cl2 = 1.41 mol Excess Reactant: S 8
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The Calculation S 8 + 4 Cl 2 4 S 2 Cl 2 1.41 mol Cl 2 1 mol S 8 256 g S 8 x x 1 4 mol Cl 2 1 mol S 8 = 90.24 g S 8
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Excess Remaining Subtract amount that reacted from the amount that you started with In example: Amount Reacted = 90.24 g S 8 Amount Starting With = 200.0 g S 8
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The Answer 200.0 g - 90.24 g = 109.76 g S 8 in excess
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