Unit 4-4 Leftover of Excess Reactant

Limiting Reactant (LR) and Excess Reactant (ER)

• LR: Reactant that gives the least amount of product and is completely consumed in the reaction.
  • There is only 1 LR per reaction.
• ER: Reactant that is not completely used up after the reaction; there is some leftover.
  • There can be >1 ER in a reaction.

Identifying LR and ER

• LR decides the number of products.
• To find LR, calculate how many moles of a product can be formed from each reactant. The reactant yielding the least product is the LR.

Calculating Leftover ER

1. Convert to Moles: Convert grams of reactants A and B to moles.
2. Find LR: Determine the limiting reactant by calculating the moles of product C that can be produced from reactants A and B.
3. Calculate Consumed ER: Convert moles of product C (actual yield) or moles of LR to moles of consumed ER.
4. Calculate Leftover ER: Subtract the moles of consumed ER from the initial moles of ER.
   • Leftover of ER = Initial ER - Consumed ER
5. Convert to Grams (if needed): Convert moles of leftover ER to grams.

Example Calculation

2 NH3(g) + 3 CuO(s) \rightarrow N2(g) + 3 Cu(s) + 3 H_2O(l)

• Given: 9.05 g NH3, 45.2 g CuO
• Molar masses: 1 mol NH3 = 17.03g, 1 mol CuO = 79.55g

Steps:

1. Convert grams to moles:
   • 9.05 g NH3 \times \frac{1 mol NH3}{17.03 g NH3} = 0.53 mol NH3
   • 45.2 g CuO \times \frac{1 mol CuO}{79.55 g CuO} = 0.57 mol CuO
2. Determine LR:
   • 0.53 mol NH3 \times \frac{1 mol N2}{2 mol NH3} = 0.265 mol N2
   • 0.57 mol CuO \times \frac{1 mol N2}{3 mol CuO} = 0.189 mol N2
   • CuO is LR, NH3 is ER
3. Calculate consumed NH3:
   • 0.189 mol N2 \times \frac{2 mol NH3}{1 mol N2} \times \frac{17.03g NH3}{1 mol NH3} = 6.44 g NH3
4. Calculate leftover NH3:
   • Leftover NH3 = 9.05g - 6.44g = 2.61 g

Alternative Method (LR to Consumed ER)

• Convert LR (CuO) to consumed ER (NH3):
  • 0.57 mol CuO \times \frac{2 mol NH3}{3 mol CuO} \times \frac{17.03 g NH3}{1 mol NH3} = 6.45 g NH3
• Leftover NH3 = 9.05 g - 6.45 g = 2.60 g

Practice Problem Example

CH4(g) + 2 O2(g) \rightarrow CO2(g) + 2 H2O(g)

• 5 moles CH4 reacts with 8 moles O2.
• Find moles of excess reactant (ER) left.
• If O2 is LR:
  • 8 moles O2 \times \frac{1 mole CH4}{2 moles O2} = 4 moles CH4
• Consumed CH4 = 4 moles
• Leftover CH4 = 5 moles - 4 moles = 1 mole