Limiting Reactant

Stoichiometry Refresher Recall: The mole ratio can be used to determine the quantity (moles) or mass of reactants consumed and products formed However, usually there is one reactant which will be consumed entirely and the other reactant is left over (in excess)

Excess Reactant: The reactant that is NOT completely consumed in a chemical reaction (i.e. it is in excess) Limiting Reactant: The reactant that is completely consumed in a chemical reaction. It limits how much product is formed.

BBQ Analogy You are having a BBQ and want to make hot dogs. You buy 2 packs of wieners and 2 packs of buns There are 10 wieners per pack and 8 buns per pack (Marketing ploy??) How many hot dogs can you make? Write the BCE 1 wiener + 1 bun  1 hot dog You can make 16 hot dogs and there will be 4 wieners left over (excess reactant) You are limited by the number of buns (limiting reactant)

Solving Limiting Reactant Problems Write the BCE Set up a stoich table Convert reactants into moles Use mole ratios from BCE to calculate the moles of product formed using each reactant The reactant that produces the least amount of product is the limiting reactant. Use this amount to do any further calculations

E. g. What mass of magnesium oxide is produced when 6 E.g. What mass of magnesium oxide is produced when 6.73 g of magnesium reacts with 8.15 g of oxygen 2 Mg(s) + O2(g)  2 MgO(s) MM 24.31 g/mol 32.00 g/mol 40.31 g/mol n = 6.73 g X 1 mol 24.31 g = 0.2769 mol = 8.15 g X 1 mol 32.00 g = 0.2547 mol Using Mg: = 0.2769 molMg X 2 molMgO 2 molMg = 0.2769 molMgO Using O2: = 0.2547 molO2 X 2 molMgO 1 molO2 = 0.5094 molMgO 0.2769 mol < 0.5094 mol  Mg limits m 6.73 g 8.15 g = 0.2769 mol X 40.31 g mol = 11.2 g

Practice! Worksheet B (no limiting reactant) Worksheet C P. 231 #1-4