1. Highland Science Department Limiting Reactants in Solution

2. Highland Science Department Limiting Reactants in Solution Limiting Reactant: the reactant that is used up first in a chemical reaction. Excess Reactant: the reactant that will still be left over when the reaction is complete

3. Highland Science Department Limiting Reactants in Solution Limiting Reactant: the reactant that is used up first in a chemical reaction. Excess Reactant: the reactant that will still be left over when the reaction is complete Finding Limiting Reactant: compare mole ratios to those needed in the balanced chemical reaction to see which one is present in lower numbers

4. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.0150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide.

5. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.0150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. G: Al 2 (SO 4 ) 3(aq) + 3Ca(OH) 2(aq)  2Al(OH) 3(s) + 3CaSO 4(aq)

6. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.0150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. G: Al 2 (SO 4 ) 3(aq) + 3Ca(OH) 2(aq)  2Al(OH) 3(s) + 3CaSO 4(aq) volume = 20.0 mL concentration = 0.0150 mol/L Al 2 (SO 4 ) 3(aq) = 0.0200 L volume = 30.0 mLconcentration = 0.0185 mol/L Ca(OH) 2(aq) = 0.0300 L M.M. Al(OH) 3(s) = 78.0 g/mol

7. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.0150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. U: mass of aluminum hydroxide

8. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.0150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. U: mass of aluminum hydroxide S: 1. limiting reactant = volume x concentration ÷ lowest # moles n VM.R.M.M. 2. concentration  moles  moles  mass L.R.L.R.Al(OH) 3 Al(OH) 3

9. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.00150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. S: 1. moles= 0.0200 L= 3.0 x 10 -4 mol Al 2 (SO 4 ) 3 moles= 0.0300 L = 5.55 x 10 -4 mol Ca(OH) 2

10. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.00150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. S: 1. moles= 0.0200 L = 3.0 x 10 -4 mol Al 2 (SO 4 ) 3 3.0 x 10 -4 mol =1equal to 1 mole moles= 0.0300 L = 5.55 x 10 -4 mol Ca(OH) 2 3.0 x 10 -4 mol =1.85less than 3 moles

11. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.00150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. S: 1. moles= 0.0200 L = 3.0 x 10 -4 mol Al 2 (SO 4 ) 3 3.0 x 10 -4 mol =1equal to 1 mole Excess moles= 0.0300 L = 5.55 x 10 -4 mol Ca(OH) 2 3.0 x 10 -4 mol =1.85less than 3 moles Limiting

12. Highland Science Department Limiting Reactants in Solution e.g. 1: Find the mass of aluminum hydroxide that precipitates when 20.0 mL of 0.00150 mol/L aqueous aluminum sulfate is mixed with 30.0 mL of 0.0185 mol/L aqueous calcium hydroxide. S: 2.mass = 0.0300 L Ca(OH) 2 Al(OH) 3 = 0.0289 g