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Stoichiometry

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11 diapositives · 991 Views · Aug 31, 2014

Stoichiometry. L imiting reagent, Excess reactant, T heoretical or Percent yield. Determining amount of product Stoichiometry is used to find the quantity of a product yielded by a reaction. For example,

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Stoichiometry
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  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 (AgNO3), the silver (Ag) would be replaced in a single displacement reaction forming aqueous copper(II) nitrate (Cu(NO3)2) and solid silver. • How much silver is produced if 16.00 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 + 2AgNO3 → Cu(NO3)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: 63.55 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 + 2AgNO3 → Cu(NO3)2 + 2Ag

  6. Step 4: • Now that the moles of Ag produced is known to be 0.5036 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 + 2AgNO3 → Cu(NO3)2 + 2Ag 54.32 g of Ag is produced when 16.00 g of Cu is added to a solution of excess silver nitrate?

  8. Additional Example: • The balanced chemical equation for propane (C3H8) reacting with excess oxygen gas (O2) is: • C3H8 + 5O2 → 3CO2 + 4H2O • What is the mass of water formed when 120 g of propane (C3H8) 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 (O2) to produce lead(II) oxide (PbO) and sulfur dioxide (SO2): • 2PbS + 3O2 → 2PbO + 2SO2

  10. To determine the theoretical yield of lead(II) oxide if 200.0 g of lead(II) sulfide and 200.0 grams of oxygen are heated in an open container: • Because a lesser amount of PbO is produced for the 200.0 g of PbS, it is clear that PbS is the limiting reagent.

  11. Percent yield is expressed in the following equation: • If 170.0 g of lead(II) oxide is obtained, then the percent yield would be calculated as follows: