Limiting Reagents. We have considered reactions with just the perfect amount of each reactant.

Limiting Reagents

Limiting Reagents

• We have considered reactions with just the perfect amount of each reactant.

Limiting Reagents


• However, reactions almost never have reactants in perfect amounts.

Limiting Reagents


• However, reactions almost never have reactants in perfect amounts.

• One reactant is completely used up and another reactant is left over.

Limiting and Excess Reactants

• The limiting reactant is the reactant that is completely used up in a chemical reaction. When the limiting reactant is used up, the reaction stops. No more new product(s) is/are being produced.



• The excess reactant is the reactant that remains after a reaction is over.



• The excess reactant is the reactant that remains after a reaction is over.

• Reactant = Reagent

Strategy

Strategy

• Determine how much product can be produced by each reactant.

Strategy

• Determine how much product can be produced by each reactant.

• The reactant that produces the least amount of product is the limiting reactant.

Example

• What is the limiting reactant when 4.80 g of Calcium are mixed with 2.00 g of Nitrogen?

Example


• Write a balanced equation.

Example


• Write a balanced equation.3Ca (s) + N2 (g) Ca3N2 (s)

Example

• Convert the masses of the reactants to moles.

Example

• Convert the masses of the reactants to moles.n = m/mm

Example

• Convert the masses of the reactants to moles.n = m/mmn = 4.80 g/40.08g·mol-1 = 0.11976 mol Can = 2.00 g/28.02g·mol-1 = 0.07138 mol N2

Example

• Find the mole ratios and determine which one to use.

Example

• Find the mole ratios and determine which one to use.Ca : N2 = 3 : 1Ca : Ca3N2 = 3 : 1N2 : Ca3N2 = 1 : 1

Example

• Use the mole ratio to determine the moles of the product from the moles of the reactant.

Example

• Use the mole ratio to determine the moles of the product from the moles of the reactant.Ca : Ca3N2 = 3 : 10.040 mol Ca3N2

N2 : Ca3N2 = 1 : 10.071 mol Ca3N2

Example

• The 0.040 mol Ca3N2 produced from Ca is less than the 0.071 mol Ca3N2 produced from N2.

Example

• The 0.040 mol Ca3N2 produced from Ca is less than the 0.071 mol Ca3N2 produced from N2.

• Therefore, the limiting reactant is Ca because the ratio between Ca and Ca3N2

results in the least amount of moles of Ca3N2 produced.

Example

• What is the mass of Calcium Nitride produced?

Example


• The reaction stops when all the Ca is used up.

Example



• The moles of Ca3N2 produced from the limiting reactant, Ca, are used to calculate the mass.

Example



• The moles of Ca3N2 produced from the limiting reactant, Ca, are used to calculate the mass.m = n(mm)

Example



• The moles of Ca3N2 produced from the limiting reactant, Ca, are used to calculate the mass.m = n(mm)m = 0.040 mol(148.26 g·mol-1) = 5.93 g

Example

• What is the mass of Nitrogen left over after the reaction has stopped?

Example


• Once all the limiting reagent, Ca, has been used up, the reaction stops.

Example


• Once all the limiting reagent, Ca, has been used up, the reaction stops.

• Once the reaction stops, there is still excess reagent, N2, left over.

Example

• Use the mole ratio to determine how much of the excess reagent reacted with all of the limiting reagent.

Example


• Ca : N2 = 3 : 1

Example


• Ca : N2 = 3 : 1

• 0.040 mol N2

Example

• Subtract the moles of N2 used up from the total moles of N2 available.

Example


• 0.071 mol – 0.040 mol = 0.031 mol

Example


• 0.071 mol – 0.040 mol = 0.031 mol• Determine the mass of N2 remaining.

Example


• 0.071 mol – 0.040 mol = 0.031 mol• Determine the mass of N2 remaining.

• m = n(mm)m = 0.031 mol(28.02 g·mol-1) = 0.88 g

Example

3Ca (s) + N2 (g) Ca3N2 (s)g

molmolmole ratio

Example

3Ca (s) + N2 (g) Ca3N2 (s)g g

mol

molmol

g

Limiting Reagents. We have considered reactions with just the perfect amount of each reactant.

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Transcript of Limiting Reagents. We have considered reactions with just the perfect amount of each reactant.