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Limiting Reactants and ICE Charts. Chemistry Cake You have 20 cups of flour, 8 cups of sugar, 30 litres of milk and 48 eggs in your kitchen. The recipe.

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Presentation on theme: "Limiting Reactants and ICE Charts. Chemistry Cake You have 20 cups of flour, 8 cups of sugar, 30 litres of milk and 48 eggs in your kitchen. The recipe."— Presentation transcript:

1 Limiting Reactants and ICE Charts

2 Chemistry Cake You have 20 cups of flour, 8 cups of sugar, 30 litres of milk and 48 eggs in your kitchen. The recipe for chemistry cake is: 3 cups of flour 2 cups of sugar 2 litres of milk +6 eggs = 1 chemistry cake

3 1.How many cakes can you make? 2.Which ingredient ran out first and limited the number of cakes you could make? 3.What and how much of each ingredient is left over? 4.What does this assignment have to do with chemistry?

4 3F + 2S + 2M + 6E → 1Cake All other ingredients are in excess 8 cups of sugar will make 4 cakes- so multiple the recipe by 4 The sugar runs out- limiting ingredient 84 2083048 0 12824 802224 4 I nitial C hange E nd

5 Limiting Reactant Problems

6 Ga + O 2 → Ga 2 O 3 14.0 mole Ga and 12.0 mole O 2 react. Find the limiting reactant, the mass of excess reactant and product made. 4 3 32 4 12.0 14.0 0 C I12.0 16.0 Cannot consume 16.0 moles – we only have 14 moles Guess that one of the reactants runs out- then check

7 Ga + O 2 → Ga 2 O 3 14.0 mole Ga and 12.0 mole O 2 react. Find the limiting reactant, the mass of excess reactant and product made. 3 4 32 4 10.5 14.0 0 C I12.0 14.0 The other reactant must run out- calculate the change 2 3 7.00 E 01.5 7.00 Subtract to get what is left- Convert back to grams x 32.0 g x 187.4 g 1 mole 48. g 1.31 x 10 3 g excesslimitingproduct

8 14.0 g of Al reacts with 94.0 g of Br 2. Find the limiting reactant, the mass of the excess reactant and product. Al + Br 2 → AlBr 3 3 2 322 14.0 g x 1 mole 94.0 g x 1 mole 27.0 g 159.8 g I0.5185 mole0.5882 mole0 mole C0.5185mole 0.7778 mole

9 14.0 g of Al reacts with 94.0 g of Br 2. Find the limiting reactant, the mass of the excess reactant and product. Al + Br 2 → AlBr 3 3 322 14.0 g x 1 mole 94.0 g x 1 mole 27.0 g 159.8 g I0.5185 mole0.5882 mole0 mole 0.5882 mole0.3921moleC 2 2 3 E0.1264 mole0.00000.3921 mole x 27.0 g x 266.7 g 1 mole 1 mole 3.41 glimiting105 g

10 25.0 g of H 3 PO 4 reacts with 94.0 g of Ca(NO 3 ) 2. Find the limiting reactant, the mass of the excess reactant and product. H 3 PO 4 + Ca(NO 3 ) 2 →Ca 3 (PO 4 ) 2 +HNO 3 3216 25.0 g x 1 mole 94.0 g x 1 mole 98.03 g 164.1 g I0.2550 mole0.5728 mole0 mole 0 mole C 0.2550 mole 3 2 0.3825 mole 1 3 0.1275 mole0.765 mole 6 1 E0 mole 0.1903 mole0.1275 mole 0.765 mole x 164.1 g x 310.3 g x 63.01 g 1 mole 1 mole 1 mole limiting = 31.2 g = 39.6 g = 48.2 g


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