# Limiting Reactants and ICE Charts. “Chemistry” Salami Sandwiches Question: If you are given one dozen loaves of bread, a gallon of mustard and three pieces.

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Limiting Reactants and ICE Charts

“Chemistry” Salami Sandwiches Question: 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? Answer: Give reasons

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:

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

1.How many cakes can you make?

2.Which ingredient ran out first and limited the number of cakes you could make?

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?

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?

Limiting Reactant Problems

In the chemical reaction: 2 Al + 3 Br 2 2 AlBr 3

In the chemical reaction: 2 Al + 3 Br 2 2 AlBr 3

In the chemical reaction: 2 Al + 3 Br 2 2 AlBr 3 2 moles of Al require 3 moles of Br 2 to produce 2 moles of AlBr 3

In the chemical reaction 2 Al + 3 Br 2 2 AlBr 3 2 moles of Al require 3 moles of Br 2 to produce 2 moles of AlBr 3 What if Al and Br 2 are not present in a perfect 2 : 3 ratio?

One reactant runs out:

Answer: One reactant runs out: There is a reactant left over :

Answer: One reactant runs out:Limiting Reactant There is a reactant left over:

Answer: One reactant runs out:Limiting Reactant There is a reactant left over: Excess Reactant

Answer: One reactant runs out:Limiting Reactant There is a reactant left over: Excess Reactant We use an ICE chart to find out how much reactant in excess is left over and how much product is produced.

Answer: One reactant runs out:Limiting Reactant There is a reactant left over:Excess Reactant We use an ICE chart to find out how much reactant in excess is left over and how much product is produced. I = Initial moles (unreacted reactants)

Answer: One reactant runs out:Limiting Reactant There is a reactant left over:Excess Reactant We use an ICE chart to find out how much reactant in excess is left over and how much product is produced. I = Initial moles (unreacted reactants) C = Change in moles (reactants consumed & products produced)

Answer: One reactant runs out:Limiting Reactant There is a reactant left over:Excess Reactant We use an ICE chart to find out how much reactant in excess is left over and how much product is produced. I = Initial moles (unreacted reactants) C = Change in moles (reactants consumed & products produced) E = End moles

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made.

2 Ca + 1 O 2 2 CaO

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I C E Write down the INITIAL moles given in the question. If grams given we need to convert to moles.

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C E

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C E Now CALCULATE the moles of each reactant needed for the reaction. Start with whichever element you want.

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C E 12.0 mol O 2

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C E 12.0mol O 2 x 2 mol Ca 1 mol O 2

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C E 12.0 mol O 2 x 2 mol Ca= 24.0 mol Ca 1 mol O 2

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C E 12.0 mol O 2 x 2 mol Ca= 24.0 mol Ca 1 mol O 2 Don’t have enough Ca! 

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C E 12.0 mol O 2 x 2 mol Ca= 24.0 mol Ca 1 mol O 2 Don’t have enough Ca !  12.0 mol Ca x 1 mol O 2 = 6.00 mol of O 2 2 mol Ca Yahooooo! Enough O 2 !!! This means for all 12 moles of Ca to be used up we need 6 moles of O 2.

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C12.0 mol 6.0 mol E

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C12.0 mol6.0 mol E Since the limiting reagent determines how much product is formed we use the moles of limiting reagent to determine the moles and mass of product

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C12.0 mol6.0 mol E 12.0mol Ca

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C12.0 mol6.0 mol E 12.0mol Ca x 2 mol CaO 2 mol Ca

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C12.0 mol6.0 mol E 12.0mol Ca x 2 mol CaO= 12.0 mol CaO 2 mol Ca

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C12.0 mol6.0 mol12.0 mol E

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C-12.0 mol-6.0 mol +12.0 mol E Now complete the ICE chart by filling in the END amounts of each species.

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C-12.0 mol-6.0 mol +12.0 mol E0 mol

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C-12.0 mol-6.0 mol +12.0 mol E0 mol6.0 mol

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C-12.0 mol-6.0 mol +12.0 mol E0 mol6.0 mol12.0 mol

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C-12.0 mol-6.0 mol +12.0 mol E0 mol6.0 mol12.0 mol Limiting Reactant

12.0 mole Ca and 12.0 mole O 2 react. Find the limiting reactant, the amount of excess reactant and the amount of product made. 2 Ca + 1 O 2 2 CaO I12.0 mol12.0 mol0 C-12.0 mol-6.0 mol +12.0 mol E06.0 mol12.0 mol Limiting ExcessReactant

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made.

2 Al + 3 Br 2 2 AlBr 3

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I C E Write down the INITIAL amounts given in the question

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C E

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C E Now CALCULATE the moles of each reactant needed for the reaction. Start with whichever element you want.

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C E 24.0 Br 2 x

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C E 24.0 Br 2 x 2 mol Al 3 mol Br 2

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C E 24.0 Br 2 x 2 mol Al= 16 mol Al 3 mol Br 2

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C E 24.0 Br 2 x 2 mol Al= 16 mol Al 3 mol Br 2 This means for all 24.0 moles of Br 2 to be used up we need 16 moles of Al.

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C16.0 mol 24.0 mol E

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C16.0 mol 24.0 mol E Since the limiting reagent determines how much product is formed we use the moles of limiting reagent to determine the moles and mass of product

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C16.0 mol 24.0 mol E 24.0 mol Br 2

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C16.0 mol 24.0 mol E 24.0 mol Br 2 x 2 mol AlBr 3 = 3 mol Br 2

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C16.0 mol 24.0 mol E 24.0 mol Br 2 x 2 mol AlBr 3 = 16 mol AlBr 3 3 mol Br 2

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C16.0 mol 24.0 mol 16.0mol E

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C-16.0 mol -24.0 mol +16.0mol E Now complete the ICE chart by filling in the END amounts of each species

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C-16.0 mol -24.0 mol +16.0mol E8.0 mol

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C-16.0 mol -24.0 mol +16.0mol E8.0 mol 0 mol

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C-16.0 mol - 24.0 mol +16.0mol E8.0 mol 0 mol 16.0mol

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C-16.0 mol -24.0 mol +16.0mol E8.0 mol 0 mol 16.0mol Excess Reactant

24.0 mole of Al reacts with 24.0 mol of Br 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. 2 Al + 3 Br 2 2 AlBr 3 I24.0 mol 24.0 mol 0 mol C-16.0 mol - 24.0 mol +16.0mol E8.0 mol 0 mol 16.0mol ExcessLimiting ReactantReactant

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made.

P 4 + 5 O 2 2 P 2 O 5

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I C E

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I C E Write down the INITIAL amounts given in the question

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol4.0mol 0mol C E

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol4.0mol 0mol C E Now CALCULATE the moles of each reactant needed for the reaction. Start with whichever element you want

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C E 14.0 mol P 4 x

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C E 14.0 mol P 4 x 5 mol O 2 1 mol P 4

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C E 14.0 mol P 4 x 5 mol O 2 = 70 mol O 2 1 mol P 4

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C E 14.0 mol P 4 x 5 mol O 2 = 70 mol O 2 1 mol P 4 That’s more O 2 than we have so redo the calculation starting with O 2

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C E 4.0 mol O 2

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C E 4.0 mol O 2 x 1 mol P 4 5.0 mol O 2

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C E 4.0 mol O 2 x 1 mol P 4 = 0.80 mol P 4 5.0 mol O 2

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C E 4.0 mol O 2 x 1 mol P 4 = 0.80 mol P 4 5.0 mol O 2 This means for all 4.0 moles of O 2 to be used up we need 0.80 moles of P 4.

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C0.80 mol 4.0 mol E

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C0.80 mol 4.0 mol E Since the limiting reagent determines how much product is formed we use the moles of limiting reagent to determine the moles and mass of product

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C0.80 mol 4.0 mol E 4.0 mol O 2 x

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C 0.80 mol 4.0 mol E 4.0 mol O 2 x 2 mol P 2 O 5 5.0 mol O 2

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C 0.80 mol 4.0 mol E 4.0 mol O 2 x 2 mol P 2 O 5 = 1.6 mol P 2 O 5 5.0 mol O 2

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C 0.80 mol 4.0 mol 1.6 mol E

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C-0.80 mol -4.0 mol +1.6 mol E Now complete the ICE chart by filling in the END amounts of each species

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C -0.80 mol -4.0 mol +1.6 mol E 13.2 mol

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C -0.80 mol -4.0 mol +1.6 mol E 13.2 mol 0 mol

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C -0.80 mol-4.0 mol +1.6 mol E13.2 mol 0 mol 1.6 mol

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C-0.80 mol-4.0 mol +1.6 mol E 13.2 mol 0 mol 1.6 mol Excess Reactant

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C-0.80 mol-4.0 mol +1.6 mol E 13.2 mol 0 mol 1.6 mol ExcessLimiting ReactantReactant

14.0 moles P 4 react with 4.0 moles O 2. Find the limiting reactant, the amount of the excess reactant and the amount of product made. P 4 + 5 O 2 2 P 2 O 5 I14.0mol 4.0mol 0mol C -0.80 mol-4.0 mol +1.6 mol E 13.2 mol 0 mol 1.6 mol ExcessLimiting ReactantReactant HOMEWORK: WORKSHEET # 6

When 79.12g of aluminum is reacted with 185.0g of hot water (a SR reaction), the products are aluminum hydroxide and hydrogen gas. Which of the reactants is in excess? How many grams of aluminum hydroxide are produced? __ Al (s) + __ H 2 O (l) __Al(OH) 3 (s) + __H 2 (g)

When 79.12g of aluminum is reacted with 185.0g of hot water (a SR reaction), the products are aluminum hydroxide and hydrogen gas. Which of the reactants is in excess? How many grams of aluminum hydroxide are produced? Write and Balance equation! 2 Al (s) + 6H 2 O (l) 2Al(OH) 3 (s) + 3H 2 (g) Mass 79.12g 185.0g 0g 0g

When 79.12g of aluminum is reacted with 185.0g of hot water (a SR reaction), the products are aluminum hydroxide and hydrogen gas. Which of the reactants is in excess? How many grams of aluminum hydroxide are produced? 2 Al (s) + 6H 2 O (l) 2Al(OH) 3 (s) + 3H 2 (g) Mass 79.12g 185.0g 0g 0g I C E Mass 79.12g Al x 1 mol Al = 2.930 mol Al 27.0 g Al 185.0g H 2 O x 1 mol H 2 O = 10.27 mol H 2 O 18.0 g H 2 O

When 79.12g of aluminum is reacted with 185.0g of hot water (a SR reaction), the products are aluminum hydroxide and hydrogen gas. Which of the reactants is in excess? How many grams of aluminum hydroxide are produced? 2 Al (s) + 6H 2 O (l) 2Al(OH) 3 (s) + 3H 2 (g) I 2.93 mol 10.27 mol 0 mol 0 mol C E Now CALCULATE the moles of each reactant needed for the reaction. Start with whichever element you want

When 79.12g of aluminum is reacted with 185.0g of hot water (a SR reaction), the products are aluminum hydroxide and hydrogen gas. Which of the reactants is in excess? How many grams of aluminum hydroxide are produced? 2 Al (s) + 6H 2 O (l) 2Al(OH) 3 (s) + 3H 2 (g) I 2.93 mol 10.27 mol 0 mol 0 mol C E 2.93 mol Al x 6 mol H 2 0 = 8.74 mol H 2 O We have enough! 2 mol Al

When 79.12g of aluminum is reacted with 185.0g of hot water (a SR reaction), the products are aluminum hydroxide and hydrogen gas. Which of the reactants is in excess? How many grams of aluminum hydroxide are produced? 2 Al (s) + 6H 2 O (l) 2Al(OH) 3 (s) + 3H 2 (g) I 2.93 mol 10.27 mol 0 mol 0 mol C E 2.93 mol Al x 6 mol H 2 0 = 8.74 mol 2 mol Al

When 79.12g of aluminum is reacted with 185.0g of hot water (a SR reaction), the products are aluminum hydroxide and hydrogen gas. Which of the reactants is in excess? How many grams of aluminum hydroxide are produced? 2 Al (s) + 6H 2 O (l) 2Al(OH) 3 (s) + 3H 2 (g) I 2.93 mol 10.27 mol 0 mol 0 mol C 2.93 mol 8.74 mol E 2.93 mol Al x 2Al(OH) 3 = 2.93 mol Al(OH) 3 2 mol Al

When 79.12g of aluminum is reacted with 185.0g of hot water (a SR reaction), the products are aluminum hydroxide and hydrogen gas. Which of the reactants is in excess? How many grams of aluminum hydroxide are produced? 2 Al (s) + 6H 2 O (l) 2Al(OH) 3 (s) + 3H 2 (g) I 2.93 mol 10.27 mol 0 mol 0 mol C -2.93 mol -8.74 mol +2.93 mol E 0 mol 1.53 mol 2.93 mol Excess

When 79.12g of aluminum is reacted with 185.0g of hot water (a SR reaction), the products are aluminum hydroxide and hydrogen gas. Which of the reactants is in excess? How many grams of aluminum hydroxide are produced? 2 Al (s) + 6H 2 O (l) 2Al(OH) 3 (s) + 3H 2 (g) I 2.93 mol 10.27 mol 0 mol 0 mol C -2.93 mol -8.74 mol +2.93 mol E 0 mol 1.53 mol 2.93 mol 2.93 mol Al(OH) 3 x 78.0 g Al(OH) 3 = 228.5 g Al(OH) 3 1 mol Al(OH) 3 Now do Worksheet # 7

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