ch 10: chemical equations & calcs
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CH 10: Chemical Equations & Calcs. Renee Y. Becker CHM 1025 Valencia Community College. What is Stoichiometry?. Chemists and chemical engineers must perform calculations based on balanced chemical reactions to predict the cost of processes. - PowerPoint PPT PresentationTRANSCRIPT
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CH 10: Chemical Equations & Calcs
Renee Y. BeckerCHM 1025
Valencia Community College
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• 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.
What is Stoichiometry?
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• Let’s look at the reaction of nitrogen monoxide with oxygen to produce nitrogen dioxide:
2 NO(g) + O2(g) → 2 NO2(g)
• Two molecules of NO gas react with one molecule of O2 gas to produce 2 molecules of NO2 gas.
Interpreting Chemical Equations
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2 NO(g) + O2(g) → 2 NO2(g)• The coefficients represent molecules, so we can multiply
each of the coefficients and look at more than individual molecules.
NO(g) O2(g) NO2(g)
2 molecules 1 molecule 2 molecules
2000 molecules 1000 molecules 2000 molecules
12.04 × 1023 molecules
6.02 × 1023 molecules
12.04 × 1023 molecules
2 moles 1 mole 2 moles
Moles & Equation Coefficients
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2 NO(g) + O2(g) → 2 NO2(g)
• We can now read the balanced chemical equation as “2 moles of NO gas react with 1 mole of O2 gas to produce 2 moles of NO2 gas.”
• The coefficients indicate the ratio of moles, or mole ratio, of reactants and products in every balanced chemical equation.
Mole Ratios
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• Recall that, according to Avogadro’s theory, there are equal number of molecules in equal volumes of gas at the same temperature and pressure.
• So, twice the number of molecules occupies twice the volume.
2 NO(g) + O2(g) → 2 NO2(g)
• So, instead of 2 molecules NO, 1 molecule O2, and 2 molecules NO2, we can write: 2 liters of NO react with 1 liter of O2 gas to produce 2 liters of NO2 gas.
Volume & Equation Coefficients
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• From a balanced chemical equation, we know how many molecules or moles of a substance react and how many moles of product(s) are produced.
• If there are gases, we know how many liters of gas react or are produced.
Interpretation of Coefficients
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• The law of conservation of mass states that mass is neither created nor destroyed during a chemical reaction. Let’s test: 2 NO(g) + O2(g) → 2 NO2(g)
– 2 mol NO + 1 mol O2 → 2 mol NO
– 2 (30.01 g) + 1 (32.00 g) → 2 (46.01 g)
– 60.02 g + 32.00 g → 92.02 g
– 92.02 g = 92.02 g
• The mass of the reactants is equal to the mass of the product! Mass is conserved.
Conservation of Mass
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• We can use a balanced chemical equation to write mole ratio, which can be used as unit factors:
N2(g) + O2(g) → 2 NO(g)
• Since 1 mol of N2 reacts with 1 mol of O2 to produce 2 mol of NO, we can write the following mole relationships:
1 mol N2
1 mol O2
1 mol N2
1 mol NO1 mol O2
1 mol NO
1 mol O2
1 mol N2
1 mol NO1 mol N2
1 mol NO1 mol O2
Mole-Mole Relationships
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• How many moles of oxygen react with 2.25 mol of nitrogen?
N2(g) + O2(g) → 2 NO(g)
Example 1
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• There are three basic types of stoichiometry problems we’ll introduce in this chapter:
–Mass-Mass stoichiometry problems
–Mass-Volume stoichiometry problems
–Volume-Volume stoichiometry problems
Types of Stoichiometry Problems
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• In a mass-mass stoichiometry problem, we will convert a given mass of a reactant or product to an unknown mass of reactant or product.
• There are three steps:
– Convert the given mass of substance to moles using the molar mass of the substance as a unit factor.
– Convert the moles of the given to moles of the unknown using the coefficients in the balanced equation.
– Convert the moles of the unknown to grams using the molar mass of the substance as a unit factor.
Mass-Mass Problems
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• What is the mass of mercury produced from the decomposition of 1.25 g of mercury(II) oxide (MM = 216.59 g/mol)?
2 HgO(s) → 2 Hg(l) + O2(g)
Example 2
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2 HgO(s) → 2 Hg(l) + O2(g)
Example 2
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• In a mass-volume stoichiometry problem, we will convert a given mass of a reactant or product to an unknown volume of reactant or product.
• There are three steps:
– Convert the given mass of a substance to moles using the molar mass of the substance as a unit factor.
– Convert the moles of the given to moles of the unknown using the coefficients in the balanced equation.
– Convert the moles of unknown to liters using the molar volume of a gas as a unit factor.
Mass-Volume Problems
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• How many liters of hydrogen are produced from the reaction of 0.165 g of aluminum metal with dilute hydrochloric acid @ STP?
2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g)
• Convert grams Al to moles Al using the molar mass of aluminum (26.98 g/mol).
• Convert moles Al to moles H2 using the balanced equation.
• Convert moles H2 to liters using the molar volume at STP.
Example 3
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2 Al(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2(g)
g Al mol Al mol H2 L H2
Example 3
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• How many grams of sodium chlorate are needed to produce 9.21 L of oxygen gas at STP?
2 NaClO3(s) → 2 NaCl(s) + 3 O2(g)
• Convert liters of O2 to moles O2, to moles NaClO3, to grams NaClO3 (106.44 g/mol).
Example 4
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• Gay-Lussac discovered that volumes of gases under similar conditions combined in small whole number ratios. This is the law of combining volumes.
• Consider the reaction: H2(g) + Cl2(g) → 2 HCl(g)
• 10 mL of H2 reacts with 10 mL of Cl2 to produce 20 mL of HCl.
• The ratio of volumes is 1:1:2, small whole numbers.
Volume-Volume Stoichiometry
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• The whole number ratio (1:1:2) is the same as the mole ratio in the balanced chemical equation:
H2(g) + Cl2(g) → 2 HCl(g)
Law of Combining Volumes
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• In a volume-volume stoichiometry problem, we will convert a given volume of a gas to an unknown volume of gaseous reactant or product.
• There is one step:
– Convert the given volume to the unknown volume using the mole ratio (therefore, the volume ratio) from the balanced chemical equation.
Volume-Volume Problems
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• How many liters of oxygen react with 37.5 L of sulfur dioxide in the production of sulfur trioxide gas?
2 SO2(g) + O2(g) → 2 SO3(g)
• From the balanced equation, 1 mol of oxygen reacts with 2 mol sulfur dioxide.
• So, 1 L of O2 reacts with 2 L of SO2.
Example 5
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2 SO2(g) + O2(g) → 2 SO3(g)
L SO2 L O2
How many L of SO3 are produced?
Example 5
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• Say you’re making grilled cheese sandwiches. You need 1 slice of cheese and 2 slices of bread to make one sandwich.
1 Cheese + 2 Bread → 1 Sandwich
• If you have 5 slices of cheese and 8 slices of bread, how many sandwiches can you make?
• You have enough bread for 4 sandwiches and enough cheese for 5 sandwiches.
• You can only make 4 sandwiches; you will run out of bread before you use all the cheese.
Limiting Reactant Concept
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• Since you run out of bread first, bread is the ingredient that limits how many sandwiches you can make.
• In a chemical reaction, the limiting reactant is the reactant that controls the amount of product you can make.
• A limiting reactant is used up before the other reactants.
• The other reactants are present in excess.
Limiting Reactant
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• 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)
• According to the balanced equation, 1 mol of Fe reacts with 1 mol of S to give 1 mol of FeS.
• So 2.50 mol of Fe will react with 2.50 mol of S to produce 2.50 mol of FeS.
• Therefore, iron is the limiting reactant and sulfur is the excess reactant.
Determining the Limiting Reactant
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• If you start with 3.00 mol of sulfur and 2.50 mol of sulfur reacts to produce FeS, you have 0.50 mol of excess sulfur (3.00 mol – 2.50 mol).
• The table below summarizes the amounts of each substance before and after the reaction.
Determining the Limiting Reactant
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There are three steps to a limiting reactant problem:
1. Calculate the mass of product that can be produced from the first reactant.
mass reactant #1 mol reactant #1 mol product mass product
2. Calculate the mass of product that can be produced from the second reactant.
mass reactant #2 mol reactant #2 mol product mass product
3. The limiting reactant is the reactant that produces the least amount of product.
Mass Limiting Reactant Problems
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• 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) + Al2O3(s)
• First, let’s convert g FeO to g Fe:
Example 6
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3 FeO(l) + 2 Al(l) → 3 Fe(l) + Al2O3(s)
• Second, lets convert g Al to g Fe:
Example 6
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• Let’s compare the two reactants:
– 25.0 g FeO can produce 19.4 g Fe
– 25.0 g Al can produce 77.6 g Fe
• FeO is the limiting reactant.
• Al is the excess reactant.
Example 6
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• When you perform a laboratory experiment, the amount of product collected is the actual yield.
• The amount of product calculated from a limiting reactant problem is the theoretical yield.
• The percent yield is the amount of the actual yield compared to the theoretical yield.
× 100 % = percent yieldactual yield
theoretical yield
Percent Yield
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• Suppose a student performs a reaction and obtains 0.875 g of CuCO3 and the theoretical yield is 0.988 g. What is the percent yield?
Cu(NO3)2(aq) + Na2CO3(aq) → CuCO3(s) + 2 NaNO3(aq)
Example 7
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• Here is a flow chart for doing stoichiometry problems.
Summary