Limiting Reactant. Identify the limiting reactant and calculate the mass of a product, given the...
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Transcript of Limiting Reactant. Identify the limiting reactant and calculate the mass of a product, given the...
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 TermsExcess 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
Limiting Reactant - determines the amount of product that can be formed in a reaction.
2 moles + 3 moles
Reactants remaining are called the excess reactants.
H
H
H
NH H
N
N
H H
H H
H
H
H
N
N2 (g) + 3 H2 (g) 2 NH3 (g)
N
N
Limiting Reactant Problems
Step 1: Record what you HAVE
Step 2: Calculate what you NEEDPick one reactant and calculate how much of the other you will need.
Step 3: Determine the limiting reactant and excess 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?
2 Na (s) + Cl2 (g) 2 NaCl (s)
1 mol Cl2
2 mol Na6.70 mol Na Cl2 (need)
Pick one reactant and calculate the other
HAVE
NEED
6.70 mol 3.20 mol
= 3.35 mol
2 Na (s) + Cl2 (g) 2 NaCl (s)
1 mol Cl2
2 mol Na3.20 mol Cl2
Both calculations lead to the same conclusion:
Cl2 - limiting reactant Na - excess reactant
Pick one reactant and calculate the other
HAVE 6.70 mol 3.20 molNEED 3.35 mol
Na (need)= 6.40 mol
= 6.40 mol NaCl1 mol Cl2
2 mol NaCl3.20 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
2 Na (s) + Cl2 (g) 2 NaCl (s)
Cl2 - limiting reactant Na - excess reactant
HAVE 6.70 mol 3.20 molNEED 6.40 mol 3.35 mol
N2 (g) + 3 H2 (g) 2 NH3 (g)
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?
1 mol N2
3 mol H2
1 mol H2
2.0 g H2
3.50 g H2 28.0 g N2
1 mole N2
= 16.3 g N2
HAVE 18.0 g 3.50 g
NEED 16.3 g
3 mol H2
1 mol N2
1 mol N2
28.0 g N2
18.0 g N2 2.02 g H2
1 mole H2
= 3.90 g H2
3.90 g
18.0 g – 16.3 g = 1.70g N2 left
N2 (g) + 3 H2 (g) 2 NH3 (g)
2 mol NH3
3 mol H2
1 mol H2
2.0 g H2
3.50 g H2 17.0 g NH3
1 mol NH3
= 19.8 g NH3
H2 - limiting reactant N2 - excess reactant
HAVE 18.0 g 3.50 g
NEED 16.3 g 3.90 g
What volume of carbon dioxide is formed by the reaction of 75.0 g of propane and 150.0 L of oxygen at STP?
HAVE 75.0 g 150.0 L
NEED 58.9 g 191 L
C3H8 (g) + O2 (g) CO2 (g) + H2O (g) 3 4 5
= 191 L O2
5 mol O2
1 mol C3H8
1 mol C3H8
44 g C3H8
75 g C3H8 22.4 L O2
1 mol O2
= 58.9 g C3H8
1 mol C3H8
5 mol O2
1 mol O2
22.4 L O2
150 L O2 44 g C3H8
1 mol C3H8
75.0 g – 58.9 g = 16.1 g C3H8 left
O2 - limiting reactant C3H8 - excess reactant
HAVE 75.0 g 150.0 L
NEED 58.9 g 191 L
C3H8 + 5 O2 3 CO2 + 4 H2O
= 90 L CO23 mol CO2
5 mol O2
1 mol O2
22.4 L O2
150 L O2 22.4 L CO2
1 mol CO2
· The limiting reactant is completely consumed. · The excess reactant is NOT used up.
When solving limiting reactant problems:
1. Balance the chemical equation first2. Find the limiting reactant3. Use limiting reactant to determine the product4. Calculate the excess
CAN YOU / HAVE YOU?
• 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 TermsExcess reactant