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Limiting Reactant

Limiting Reactant. Chem 332 – O’Dette. Why care about the limiting reactant?. Limiting reactant (LR) tells you the max amount of product that can be produced LR is always completely used up ** Remember the smores lab, what was our limiting reactant?!?!.

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Limiting Reactant

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  1. Limiting Reactant Chem 332 – O’Dette

  2. Why care about the limiting reactant? Limiting reactant (LR) tells you the max amount of product that can be produced LR is always completely used up ** Remember the smores lab, what was our limiting reactant?!?!

  3. What happens to the amount of reactant that is present in excess? Called the excess reactant (ER) The ER will not react and be left over

  4. Steps to Calculating Product from the LR Find the number of moles of both reactants Determine the number of moles of one product made from each reactant Identify the LR by the reactant that produces the smaller amount of moles for the product Use the LR to find the mass of the product

  5. Example #1 Ethane undergoes the following combustion reaction: C2H4 + 3 O2 2 H2O + 2 CO2 If 2.70 moles C2H4 is reacted with 6.30 mole of O2, identify the limiting reactant and determine the number of moles of carbon dioxide produced.

  6. Example #1 (continue) 5.4 mole CO2 1 mole C2H4 4.2 mole CO2 3 mole O2 LR = O2 C2H4 + 3 O2 2 H2O + 2 CO2 2.70 mole C2H4 x 2 moles CO2 = 6.30 mole O2 x 2 moles CO2 =

  7. Example #2 Hydrogen gas can be produced by the reaction of magnesium metal with hydrochloric acid. Identify the limiting reactant when 6.00 g of HCl reacts with 5.00 g of Mg and determine the mass of hydrogen produced. Mg + 2HCl MgCl2 + H2

  8. Example #2 1 mole HCl 1 mole H2 x = 0.0823 mole H2 2 mole HCl 36.46 g HCl LR = HCl 1 mole Mg 1 mole H2 x 0.206 mole H2 = 1 mole Mg 24.31 g Mg 2.02 g H2 0.166 g H2 = 1 mole H2 Mg + 2 HCl MgCl2 + H2 6.00 g HCl x 5.00 g Mg x 0.0823 moles H2 x

  9. Example #3 What is the maximum number of grams of Cu2S that can be formed when 80.0 g of Cu reacts with 25.0 g S? 2 Cu + S Cu2S

  10. Example #3 1 mole Cu 1 mole Cu2S x = 0.629 mole Cu2S 2 mole Cu 63.55 g Cu LR = Cu 1 mole S 1 mole Cu2S x 0.780 mole Cu2S = 1 mole S 32.07 g S 159.17 g Cu2S 100. g Cu2S = 1 mole Cu2S 2 Cu + S Cu2S 80.0 g Cu x 25.0 g S x 0.629 moles Cu2S x

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