Limiting Reactant
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Limiting Reactant. Identify the limiting reactant and calculate the mass of a product, given the reaction equation and reactant data. Include: theoretical yield , experimental yield Additional KEY Terms Excess reactant.

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  • Identify the limiting reactantand calculate the mass of a product, given the reaction equation and reactant data.

  • Include: theoretical yield, experimental yield

  • Additional KEY Terms

  • Excess reactant


How many BLT Sandwiches can you make with 3 Tomatoes, 2 Kg of Bacon, 1 head of lettuce and 4 slices of bread?

2


N2(g) + 3 H2(g)

2 NH3(g)

H

H

H

H

N

N

N

H

N

H

H

H

N

N

H

H

H

H

Limiting Reactant- determinesthe amount of product that can be formed in a reaction.

2 moles + 3 moles

Reactants remaining are called the excess reactants.


Limiting Reactant Problems

Step 1: Record what you HAVE

Step 2: Calculatewhat you NEED

Pick one reactant and calculate how much of the other you will need.

Step 3: Identify the limiting reactant

Step 4: Use limiting reactant to determine the amount of product.


2 Na (s) + Cl2 (g) 2 NaCl (s)

How much NaCl is produced and what is left over when 6.70 mol of Na react with 3.20 mol of Cl2?

6.70 mol

3.20 mol

HAVE

NEED

Pick one reactant and calculate the other

6.70 mol Na

1 mol Cl2

=

3.35 mol

Cl2 (need)

2 mol Na


2 Na (s) + Cl2 (g) 2 NaCl (s)

HAVE 6.70 mol 3.20 mol

NEED 3.35 mol

Pick one reactant and calculate the other

3.20 mol Cl2

2 mol Na

=

6.40 mol

Na(need)

1 mol Cl2

Both calculations lead to the same conclusion:

Have too much Na and Don’t have enough Cl2


2 Na (s) Cl2 (g) 2 NaCl(s)

HAVE 6.70 mol 3.20 mol

-

NEED 6.40 mol 3.35 mol

Na - excess reactant Cl2- limiting reactant

3.20 mol Cl2

2 mol NaCl

= 6.40 mol NaCl

1 mol Cl2

You could use your data to calculate exactly how much excess is left over:

6.70 mol - 6.40 mol = 0.30 mol Na excess


How many grams of ammonia can be made from 3.50 g of H2 gas and 18.0 g of nitrogen gas? What’s left?

N2(g) + 3 H2 (g) 2 NH3(g)

HAVE 18.0 g 3.50 g

3.90 g

NEED

16.3 g

3.50 g H2

1 mol H2

1 mol N2

28.0 g N2

= 16.3 g N2

3 mol H2

2.0 g H2

1 mole N2

18.0 g N2

1 mol N2

3 mol H2

2.02 g H2

= 3.90 g H2

1 mol N2

28.0 g N2

1 mole H2


N2(g) + 3 H2 (g) 2 NH3(g)

HAVE 18.0 g 3.50 g

3.90 g

NEED

16.3 g

N2 - excess reactant H2- limiting reactant

3.50 g H2

17.0 g NH3

1 mol H2

2 mol NH3

1 mol NH3

3 mol H2

2.0 g H2

= 19.8 g NH3

18.0 g – 16.3 g = 1.70g N2 left


What volume of carbon dioxide is formed by the reaction of 75.0 g of propane and 150.0 L of oxygen at STP?

C3H8(g) + O2 (g) CO2 (g) + H2O (g)

5

3

4

HAVE 75.0 g 150.0 L

191 L

NEED

58.9 g

75 g C3H8

22.4 L O2

1 mol C3H8

5 mol O2

= 191 L O2

1 mol O2

1 mol C3H8

44 g C3H8

150 L O2

44 g C3H8

1 mol O2

1 mol C3H8

= 58.9 g C3H8

5 mol O2

22.4 L O2

1 mol C3H8


C3H8 + 5 O2 3 CO2 + 4 H2O

HAVE 75.0 g 150.0 L

191 L

NEED

58.9 g

C3H8- excess reactant O2 - limiting reactant

150 L O2

1 mol O2

3 mol CO2

22.4 L CO2

= 90 L CO2

5 mol O2

1 mol CO2

22.4 L O2

75.0 g – 58.9 g = 16.1 g C3H8 left


  • The limiting reactant is completely consumed.

  • The excess reactant is NOT used up.

  • When solving limiting reactant problems:

  • Balance the chemical equation first

  • Find the limiting reactant

  • Use limiting reactant to determine the product

  • Calculate the excess


  • CAN YOU / HAVE YOU?

  • Identifythe limiting reactantand calculate the mass of a product, given the reaction equation and reactant data.

  • Include: theoretical yield, experimental yield

  • Additional KEY Terms

  • Excess reactant


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