Unit 7
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Unit 7. Limiting Reactants. Limiting Reactants. Limiting Reactants. N 2 (g) + 3 H 2 (g)  2 NH 3 (g) If we start off with 100 moles of N 2 and 100 moles of H 2 , how many moles of NH 3 can be produced?. Limiting Reactants. N 2 (g) + 3 H 2 (g)  2 NH 3 (g)

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Unit 7

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Unit 7

Unit 7

Limiting Reactants


Limiting reactants

Limiting Reactants


Limiting reactants1

Limiting Reactants

N2 (g) + 3 H2 (g)  2 NH3 (g)

If we start off with 100 moles of N2 and 100 moles of H2, how many moles of NH3 can be produced?


Limiting reactants2

Limiting Reactants

N2 (g) + 3 H2 (g)  2 NH3 (g)

If we start off with 100 moles of N2 and 100 moles of H2:


Limiting reactants3

Limiting Reactants

N2 (g) + 3 H2 (g)  2 NH3 (g)

If we start off with 100 moles of N2 and 100 moles of H2:


Limiting reactants4

Limiting Reactants

N2 (g) + 3 H2 (g)  2 NH3 (g)

If we start off with 100 moles of N2 and 100 moles of H2:


Limiting reactants5

Limiting Reactants

N2 (g) + 3 H2 (g)  2 NH3 (g)

If we start off with 100 moles of N2 and 100 moles of H2, how many moles of NH3 can be produced?

How much reactants are left?


Limiting reactants6

Limiting Reactants

Cu(s) + H2SO4(aq)  CuSO4(aq) + H2O(l) + SO2(g)

How much SO2 (in grams) can be produced from 18.00 g of Cu and 18.00 g of H2SO4?

How much copper is left over?


Limiting reactants7

Limiting Reactants

Aluminum oxide can be produced from the following chemical reaction

How much Al2O3 can be prepared (in grams) from50.0 g of Al and 100.0 g of O2?

4 Al + 3 O2 2 Al2O3

10.0 g of Al and 10.0 g of O2

11.0 g of Al and 9.40 g of O2


Theoretical and percent yield

Theoretical and Percent Yield

Theoretical Yield

Maximum amount of a product that can be produced from a given amount of reactants

Actual Yield

The amount of a product actually obtained from a chemical reaction

Percent Yield = Actual Yield

Theoretical Yield

4 Al + 3 O2 2 Al2O3

If the Percent Yield is 82%, how much Al2O3 is actually produced?

50.0 g of Al and 100.0 g of O2


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