# Limiting Reactant.

## Presentation on theme: "Limiting Reactant."— Presentation transcript:

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

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.
N2 (g) H2 (g) 2 NH3 (g) 2 moles moles N H H N H N N H H Reactants remaining are called the excess reactants.

Limiting Reactant Problems
Step 1: Record what you HAVE Step 2: Calculate what you NEED Pick 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.

Pick one reactant and calculate the other
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 2 Na (s) + Cl2 (g) NaCl (s) 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) = 6.40 mol Na (need) 3.20 mol Cl2
HAVE mol mol NEED 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: Cl2 - limiting reactant Na - excess reactant

Cl2 - limiting reactant Na - excess reactant
2 Na (s) + Cl2 (g) NaCl (s) HAVE mol mol NEED mol mol Cl2 - limiting reactant Na - excess reactant 3.20 mol Cl2 2 mol NaCl = mol NaCl 1 mol Cl2 You could use your data to calculate exactly how much excess is left over: 6.70 mol mol = 0.30 mol Na excess

N2 (g) + 3 H2 (g) 2 NH3 (g) = 16.3 g N2 = 3.90 g H2
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) H2 (g) NH3 (g) HAVE g g NEED 16.3 g 3.90 g 3.50 g H2 1 mol H2 1 mol N2 28.0 g N2 = g N2 2.0 g H2 3 mol H2 1 mole N2 18.0 g N2 1 mol N2 3 mol H2 2.02 g H2 = g H2 28.0 g N2 1 mol N2 1 mole H2

H2 - limiting reactant N2 - excess reactant
N2 (g) H2 (g) NH3 (g) HAVE g g NEED 16.3 g 3.90 g H2 - limiting reactant N2 - excess reactant 3.50 g H2 1 mol H2 2 mol NH3 17.0 g NH3 2.0 g H2 3 mol H2 1 mol NH3 = g NH3 18.0 g – 16.3 g = 1.70g N2 left

5 3 4 C3H8 (g) + O2 (g) CO2 (g) + H2O (g)
What volume of carbon dioxide is formed by the reaction of 75.0 g of propane and L of oxygen at STP? C3H8 (g) O2 (g) CO2 (g) + H2O (g) 5 3 4 HAVE g L NEED 58.9 g 191 L 75 g C3H8 1 mol C3H8 5 mol O2 22.4 L O2 = 191 L O2 44 g C3H8 1 mol C3H8 1 mol O2 150 L O2 1 mol O2 1 mol C3H8 44 g C3H8 = 58.9 g C3H8 22.4 L O2 5 mol O2 1 mol C3H8

O2 - limiting reactant C3H8 - excess reactant
C3H O CO H2O HAVE g L NEED 58.9 g 191 L O2 - limiting reactant C3H8 - excess reactant 150 L O2 1 mol O2 3 mol CO2 22.4 L CO2 = 90 L CO2 22.4 L O2 5 mol O2 1 mol CO2 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? 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