chapter 12
DESCRIPTION
Chapter 12. Stoichiometry. Stoichiometry. STOY-KEE-AHM-EH-TREE Founded by Jeremias Richter, a German chemist Greek orgin: stoikheion: element & metron: measure Stoichiometry —the calculation of quantities in chemical reactions - PowerPoint PPT PresentationTRANSCRIPT
Stoichiometry STOY-KEE-AHM-EH-TREE
Founded by Jeremias Richter, a German chemist Greek orgin: stoikheion: element & metron: measure
Stoichiometry—the calculation of quantities in chemical reactions Calculating how much of each reactant is
needed and how much of each product is formed.
Like a recipe: you need a certain amount of ingredients to get a certain amount of product.
Chemical Equations: What can you tell?
1N2 + 3H2 2NH3The coefficients in a Balanced Chemical
Equation tell the # of particles of each reactant & product
1 molecule of N2 reacts with 3 molecules of H2 to produce 2 molecules of NH3.
OR The coefficients can represent the # of moles of each reactant & product.
1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
1. You have to use units to understand the process
2. Units include g, mol, L, particles & chemical formulas
3. Convert your given to moles
4. You have to use units to understand the process
5. Use BCE coefficients to convert from one substance to another
6. You have to use units to understand the process
7. Convert answer to desired unit
8. You have to use units to understand the process
9. Units include g, mol, L, particles & chemical formulas
10. It really is not that hard
You Should Know About Stoichiometry
3 Steps of Stoichiometry1. Convert to moles (from particles, mass, or
volume).
2. Do a mole to mole conversion using the balanced chemical equation
3. Convert moles to desired unit (particles, mass, or volume).
1 Mole 1 MoleBCE
1. 2. 3.
Stoichiometry Example
1N2 + 3H2 2NH3
If 2.61 L of H2 gas reacts with the nitrogen in the air, how many grams of NH3 will be produced?
2.61 L H2 x 1 mol H2_22.4 L H2
2 mol NH3 3 mol H2
17.0 g NH3_ 1 mol NH3
= 1.32 g NH3
xx
Given 1 Mole BCE 1 Mole
Percent Yield In a lab, the amount of product formed is often less than expected based on the balanced chemical equation.
Theoretical Yield—the amount of product that should be formed (from stoichiometry problem)
Actual Yield—the amount of product actually formed in a lab (found in an experiment).
Percent Yield—the ratio of actual to theoretical yield. Percent Yield = actual yield x 100
theoretical yield
Example 24.8g calcium carbonate is decomposed by heating.
13.1g CaO is actually produced. What is the percent yield?
CaCO3 CaO + CO2
Actual Yield: 13.1g CaO Theoretical Yield: (use stoichiometry)
24.8g CaCO3 x 1 mol CaCO3 x 1 mol CaO x 56.1 g CaO
100.1 g CaCO3 1 mol CaCO3 1 mol CaO
=13.9 g CaO % Yield = 13.1 x100
13.9
Limiting Reagents
Limiting Reagent—limits or determines the amount of product that can be formed in a reaction.
Excess Reagent—the reactant that is not completely used up in a reaction.
Limiting Reactants video
5N2 and 9H2 6NH3 and 2N2
Hydrogen is Oxygen is limiting excess
Hydrogen islimiting
NitrogenNitrogen is excess
Solving for a Limiting Reagent:
1.1. Use stoichiometry to convert each Use stoichiometry to convert each reactant (individually) into product reactant (individually) into product (that’s 2 stoichiometry problems)(that’s 2 stoichiometry problems)
2.2. The reactant that gives you the least The reactant that gives you the least product is the product is the limitinglimiting reagent. reagent.
3.3. The reactant that gives the most The reactant that gives the most product is the product is the excess excess reagent.reagent.
Example: Nickel replaces silver from silver nitrate in solution according to the
following equation: 2AgNO3 + Ni → 2Ag + Ni(NO3)2 If you have 22.9 g of Ni and 112 g of AgNO3 ,what mass of nickel(II) nitrate would be produced?
22.9g Ni x 1 mol Ni x 1 mol Ni(NO3)2 x 182.7g Ni(NO3)2 = 71.3 g Ni(NO3)2
58.7g Ni 1 mol Ni 1 mol Ni(NO3)2
112 g AgNO3x1 mol AgNO3 x 1 mol Ni(NO3)2 x 182.7g Ni(NO3)2 =60.2g Ni(NO3)2
169.9g AgNO3 2 mol AgNO3 1 mol Ni(NO3)2
AgNO3 is limiting & 60.2 g Ni(NO3)2 can be produced