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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

Limiting Reactants

### 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 Reactants

N2 (g) + 3 H2 (g)  2 NH3 (g)

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

### 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 Reactants

N2 (g) + 3 H2 (g)  2 NH3 (g)

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

### 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 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 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 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