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