Chemical reactions

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Rxn 2: Balloon inflates fully, no Zn left

* Right amount of each (HCl and Zn)

Rxn 3: Balloon does not inflate fully, no Zn left.

* Not enough Zn to use up 0.100 mol HCl

LIMITING REACTANTS

React solid Zn with 0.100 mol HCl (aq)

Zn + 2 HCl ---> ZnCl2 + H2

(See CD Screen 4.8)

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Rxn 1 Rxn 2 Rxn 3

mass Zn (g) 7.00 3.27 1.31

mol Zn 0.107 0.050 0.020

mol HCl 0.100 0.100 0.100

mol HCl/mol Zn 0.93/1 2.00/1 5.00/1

Lim Reactant LR = HCl no LR LR = Zn

LIMITING REACTANTS

React solid Zn with 0.100 mol HCl (aq)

Zn + 2 HCl ---> ZnCl2 + H2

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Reaction to be Studied

2 Al + 3 Cl2 ---> Al2Cl6

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PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?

Stoichiometric

factor

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Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

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2 Al + 3 Cl2 ---> Al2Cl6

Reactants must be in the mole ratio

Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

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Deciding on the Limiting Reactant

If

There is not enough Al to use up all the Cl2

2 Al + 3 Cl2 ---> Al2Cl6

Lim reag = Al

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If

There is not enough Cl2 to use up all the Al

2 Al + 3 Cl2 ---> Al2Cl6

Lim reag = Cl2

Deciding on the Limiting Reactant

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We have 5.40 g of Al and 8.10 g of Cl2

Step 2 of LR problem: Calculate moles of each reactant

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Find mole ratio of reactants

Limiting reagent is Cl2

2 Al + 3 Cl2 ---> Al2Cl6

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Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?

Limiting reactant = Cl2

Base all calcs. on Cl2

2 Al + 3 Cl2 ---> Al2Cl6

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CALCULATIONS: calculate mass of

Al2Cl6 expected.

Step 1: Calculate moles of Al2Cl6 expected based on LR.

Step 2: Calculate mass of Al2Cl6 expected based on LR.

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