Limiting Reagents Notes

Limiting Reagents

  • A reagent is another term for a reactant.

Limiting & Excess Reactants

  • A limiting reactant determines the extent of a reaction and the amount of product formed.
  • The reaction stops when the limiting reactant is completely used up.
  • Excess reactants are the reactants that remain after the limiting reactant is used up.

Product Calculation

  • To calculate the amount of product formed, use the ratios from the balanced chemical equation.
  • Identify the limiting reactant to determine the maximum amount of product that can be formed.

Example 1: Finding the Limiting Reactant & Product Amount

Problem: How much MgOMgO is produced when 48.6g48.6g of MgMg react with 64.0g64.0g of O2O_2?

Mg+O2MgOMg + O_2 \rightarrow MgO

  1. Balance the chemical equation:
    2Mg+O22MgO2Mg + O_2 \rightarrow 2MgO
  2. Find the moles of both reactants:
    • 48.6gMg×1molMg24.31g=2.00molMg48.6 g Mg \times \frac{1 mol Mg}{24.31 g} = 2.00 mol Mg
    • 64.0gO<em>2×1molO</em>232g=2.00molO264.0 g O<em>2 \times \frac{1 mol O</em>2}{32 g} = 2.00 mol O_2
  3. Multiply the moles of the reactants by mole ratios to compare the amounts of product each can produce:
    • 2.00molMg×2molMgO2molMg=2.00molMgO2.00 mol Mg \times \frac{2 mol MgO}{2 mol Mg} = 2.00 mol MgO
    • 2.00molO<em>2×2molMgO1molO</em>2=4.00molMgO2.00 mol O<em>2 \times \frac{2 mol MgO}{1 mol O</em>2} = 4.00 mol MgO
  • Magnesium produces fewer moles of magnesium oxide than oxygen, so magnesium is the limiting reactant.
  1. Convert moles of limiting reactant to grams of product to find the maximum amount of product formed:
    2.00molMgO×40.31g1molMgO=80.6gMgO2.00 mol MgO \times \frac{40.31g}{1 mol MgO} = 80.6g MgO
  • The maximum amount of MgOMgO that can be made before the limiting reactant runs out is 80.6g80.6 g.