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Stoichiometry Limiting reagent, Excess reactant, Theoretical or Percent yield.

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Presentation on theme: "Stoichiometry Limiting reagent, Excess reactant, Theoretical or Percent yield."— Presentation transcript:

1 Stoichiometry Limiting reagent, Excess reactant, Theoretical or Percent yield

2 Determining amount of product  Stoichiometry is used to find the quantity of a product yielded by a reaction. For example,  If a piece of solid copper (Cu) were added to an aqueous solution of silver nitrate (AgNO 3 ), the silver (Ag) would be replaced in a single displacement reaction forming aqueous copper(II) nitrate (Cu(NO 3 ) 2 ) and solid silver.  How much silver is produced if grams of Cu is added to the solution of excess silver nitrate?

3 The following steps are used:  Step 1 - Write and Balance the Equation  Step 2 - Mass to Mole: Convert g Cu to moles Cu  Step 3 - Mole Ratio: Convert moles of Cu to moles of Ag produced  Step 4 - Mole to Mass: Convert moles Ag to grams of Ag produced

4 Step 1:  The complete balanced equation would be: Cu + 2AgNO 3 → Cu(NO 3 ) 2 + 2Ag Step 2:  For the mass to mole step, the amount of copper (16.00 g) would be converted to moles of copper by dividing the mass of copper by its molecular mass: g/mol.

5 Step 3:  Now that the amount of Cu in moles (0.2518) is found, we can set up the mole ratio. This is found by looking at the coefficients in the balanced equation: Cu and Ag are in a 1:2 ratio. Cu + 2AgNO 3 → Cu(NO 3 ) 2 + 2Ag

6 Step 4:  Now that the moles of Ag produced is known to be mol, we convert this amount to grams of Ag produced to come to the final answer:

7  This set of calculations can be condensed into a single step: Cu + 2AgNO 3 → Cu(NO 3 ) 2 + 2Ag g of Ag is produced when g of Cu is added to a solution of excess silver nitrate?

8 Additional Example:  The balanced chemical equation for propane (C 3 H 8 ) reacting with excess oxygen gas (O 2 ) is: C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O  What is the mass of water formed when 120 g of propane (C 3 H 8 ) is burned in excess oxygen is:

9  The limiting reagent is the reagent that limits the amount of product that can be formed and is completely consumed during the reaction.  The excess reactant is the reactant that is left over once the reaction has stopped due to the limiting reactant.  Consider the equation of roasting lead(II) sulfide (PbS) in oxygen (O 2 ) to produce lead(II) oxide (PbO) and sulfur dioxide (SO 2 ): 2PbS + 3O 2 → 2PbO + 2SO 2

10  To determine the theoretical yield of lead(II) oxide if g of lead(II) sulfide and grams of oxygen are heated in an open container:  Because a lesser amount of PbO is produced for the g of PbS, it is clear that PbS is the limiting reagent.

11 Percent yield is expressed in the following equation:  If g of lead(II) oxide is obtained, then the percent yield would be calculated as follows:


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