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Chem 1075 Chapter 10 Stoichiometry Lecture Notes
Slide 2 What is stoichiometry?
- Chemists and chemical engineers must perform calculations based on balanced chemical reactions to predict the cost of processes.
- These calculations are used to avoid using large excess amounts of costly chemicals.
- The calculations these scientists use are called stoichiometry calculations.
Slide 3 Interpreting Chemical Equations
2 NO(g) + O 2 (g) à 2 NO 2 (g)
Slide 4 Moles and Equation Coefficients
2 NO(g) + O 2 (g) à 2 NO 2 (g)
NO (g) O2(g) NO2(g)
molecules molecule molecules
molecules molecules molecules
molecules molecules molecules
moles mole moles
Slide 5 Mole Ratios (Coefficient Ratios)
2 NO(g) + O 2 (g) à 2 NO 2 (g)
- We can now read the balanced chemical equation as “_______________________________ gas react with _________________________________ gas to produce ________________________ gas”.
- The coefficients indicate the ________________________________, or the ratio of the moles, of reactants and products in every balanced chemical equation.
- This mole ratio is also called the _________________________________________
Slide 6 Interpretation of Coefficients
- From a balanced chemical equation, we know how many _________________________________ ___________________of a substance react and how many ______________________________ ___________________of product(s) are produced.
- If there are gases, we know how many ________________of gas react or are produced.
Slide 7 Conservation of Mass
- The law of conservation of mass states that mass is neither created nor destroyed during a chemical reaction. Lets test: 2 NO(g) + O 2 (g) → 2 NO 2 (g)
Slide 8 Mole – Mole Relationships
- We can relate ___________________________________________to ______________________ ____________________________________ in a reaction by using coefficients (CR = coefficient relationship) from the balanced equation.
Slide 9 Mole – Mole Relationships
- Use the balanced chemical equation to write “CR” which can be used as unit factors: N 2 (g) + O 2 (g) → 2 NO(g)
- Since 1 mol of N 2 reacts with 1 mol of O 2 to produce 2 mol of NO, we can write the following mole relationships:
Slide 10 Mole – Mole Calculations
- How many moles of NO will be formed when 2.25 mol of nitrogen react? N 2 (g) + O 2 (g) → 2 NO(g)
- Since you run out of bread first, bread is the ingredient that _____________________ how many sandwiches you can make.
- In a chemical reaction, the__________________________________________is the reactant that controls the amount of products you can make.
- A limiting reactant is __________________ before the other reactants.
- The other reactants are present in excess and are referred to as ____________________________.
Slide 17-18 Determining the Limiting Reactant
- If you heat 2.50 mol of Fe and 3.00 mol of S, how many moles of FeS are formed? Fe(s) + S(s) → FeS(s)
Slide 19 Mass Limiting Reactant Problems
There are three steps to a limiting reactant problem:
- Calculate the _________________________________ that can be produced from the __________ reactant. mass reactant #1 ⇒ mol reactant #1 ⇒ mol product ⇒ mass product
- Calculate the _________________________________ that can be produced from the___________ reactant. mass reactant #2 ⇒ mol reactant #2 ⇒ mol product ⇒ mass product
- The limiting reactant is the reactant that produces the_____________________amount of product.
Slide 20-22 Mass Limiting Reactant Problem
- How much molten iron is formed from the reaction of 25.0 g FeO and 25.0 g Al? 3 FeO(l) + 2 Al(l) → 3 Fe(l) + Al 2 O 3 (s)
Slide 23 Percent Yield
- When you perform a ___________________________________________, the amount of product collected is the______________________________.
- The amount of product calculated from a limiting reactant problem is the ____________________ ___________________.
- The ______________________ is the amount of the actual yield compared to the theoretical yield.
Slide 24 Calculating Percent Yield
- Suppose a student performs a reaction and obtains 0.875 g of CuCO 3 and the theoretical yield is 0.988 g. What is the percent yield? Cu(NO 3 ) 2 (aq) + Na 2 CO 3 (aq) → CuCO 3 (s) + 2 NaNO 3 (aq)
