Limiting Reactant Calculations What happens if you don’t mix chemicals in the exact ratio of the recipe?

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Presentation transcript:

Limiting Reactant Calculations What happens if you don’t mix chemicals in the exact ratio of the recipe?

FeCl NaOH → Fe(OH) NaCl Does adding more NaOH to a constant mass of FeCl 3 always produce more product? All FeCl 3 is used up; adding more NaOH does not make any more product

Recipe for a ice cream fudge sundae: 2 scoops + 2 spoons + 1 cherry How many sundae’s could be made by doubling ice cream?

1 Sundae – need to double amounts of chocolate syrup and cherries also 2 scoops of ice cream left over – out of syrup and cherries! Amount of syrup and cherries LIMITS # of Sundaes made

Limiting Reactant - Pictures Mg + I 2 → MgI 2 Mg = I = A) 1 atom of Mg + 2 molecules I 2 → molecule(s) of MgI 2 Limiting: Excess: 1 Mg

Limiting Reactant - Pictures Mg + I 2 → MgI 2 Mg = I = B) 3 atom of Mg + 2 molecules I 2 → molecule(s) of MgI 2 Limiting: Excess: 2 I2I2

X + Y → Z If stoichiometry of reaction is 1:1: If X = Y, no limiting reactant If X > Y, Y is limiting (determines amount of Z) If X<Y, X is limiting (determines amount of Z) If stoichiometry between X and Y is not 1:1, then must account for reacting ratio (can’t simple compare amounts)

Limiting Reactant - Pictures 2 H 2 + O 2 → 2 H 2 O H = O = A) 2 molecules of H molecules O 2 → molecule(s) of H 2 O Limiting: Excess: 2 H2H2 Link to CU Phet Reactant Leftovers

Recipe for a ice cream fudge sundae: 2 scoops + 2 spoons + 1 cherry How many sundae’s could be made from

Amt ice cream 2 sundaes Amt chocolate syrup and cherries 1 sundae Only 1 sundae can actually be made; Out of syrup and cherries

2 N 2 H 4 + N 2 O 4 → 3 N H 2 O How many grams of N 2 gas are formed when 100. g of N 2 H 4 and 200. g of N 2 O 4 are mixed? Strategy: In 2 Separate calculations, determine amounts of N 2 that could be formed when each reactant is consumed. Compare amounts formed, smaller amount is actual answer (When limiting reactant runs out reaction stops) g? g 200. g 100 g N 2 H 4 → X g N g N 2 H 4 → Y g N 2 Small # of X vs Y is true theoretical yield

2 N 2 H 4 + N 2 O 4 → 3 N H 2 O How many grams of N 2 gas are formed when 100. g of N 2 H 4 and 200. g of N 2 O 4 are mixed? g N 2 H 4 ( g N 2 H 4 ) ( 3 mole N 2 ) 1 mole N 2 ( g N 2 ) = 131 g N g? g 1 mole N 2 H 4 2 mole N 2 H g

2 N 2 H 4 + N 2 O 4 → 3 N H 2 O How many grams of N 2 gas are formed when 100. g of N 2 H 4 and 200. g of N 2 O 4 are mixed? g N 2 O 4 ( g N 2 O 4 ) ( 3 mole N 2 ) 1 mole N 2 ( g N 2 ) = 183 g N g? g 1 mole N 2 O g

2 N 2 H 4 + N 2 O 4 → 3 N H 2 O How many grams of N 2 gas are formed when 100. g of N 2 H 4 and 200. g of N 2 O 4 are mixed? 100. g of N 2 H g N g of N 2 O g N 2 Final Answer = 131 g N 2 (smaller #) because N 2 H 4 runs out before N 2 O 4 stopping the reaction g? g 200. g