Post on 15-Dec-2015
Learning Goal 4: Use molar relationships in a balanced chemical reaction to predict the mass of product produced in a simple chemical reaction that goes to completion.
Substance Symbol Unit Mass
Graham Cracker S 7.0 g
Marshmallow Mm 7.1 g
Chocolate Pieces Or 3.3 g
S’more S2MmOr3 __________ g
2S + Mm + 3Or S2MmOr3The equation above gives the recipe
to make one S’more.
The equation for a chemical reaction gives the ingredients needed to make a specific amount of product.
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9-3
Table 9.1
Calculating Masses of Reactants Calculating Masses of Reactants and Productsand Products
1.1. Balance the equationBalance the equation..
2.2. Convert mass to moles. Convert mass to moles.
3.3. Set up mole ratios. Set up mole ratios.
4.4. Use mole ratios to calculate Use mole ratios to calculate moles of desired molecule. moles of desired molecule.
5.5. Convert moles to grams or to Convert moles to grams or to particles, if necessary.particles, if necessary.
Working a Stoichiometry Working a Stoichiometry ProblemProblem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.
Identify reactants and products and write the balanced equation.
4Al + 3O2 2Al2O3
What are the reactants?
Every reaction needs a yield sign!
What are the products?
What are the balanced coefficients?
Working a Stoichiometry ProblemWorking a Stoichiometry Problem
6.50 grams of aluminum reacts with an excessof oxygen. How many grams of aluminum oxide are formed?
4 Al + 3 O2 2Al2O3
=6.50 g Al
? g Al2O3
1 mol Al
26.98 g Al 4 mol Al
2 mol Al2O3
1 mol Al2O3
101.96 g Al2O3
6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 =
12.3 g Al2O3
Like all percent problems:
part
whole
1) Find the mass of each of the components (the elements),
2) Next, divide by the total mass of the compound; then x 100
x 100 % = percent
ExampleCalculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S.
29.0 g Ag33.3 g total
X 100 = 87.1 % Ag
4.30 g S33.3 g total
X 100 = 12.9 % S
Total = 100 %
Theoretical Yield The amount of product predicted from the
amount of reactant used.
The maximum amount of product that could form.
Actual Yield The amount of product that is actually produced.
Percent Yield
Actual Yield X 100% = Percent Yield
Theoretical Yield
First write the balanced equation.
2H2 + CO CH3OH
Next Calculate moles of reactants.
68,500 g CO 1 Mol CO = 2450 mol CO
28.01 g CO
8600 g H2 1 Mol H2 = 4270 mol H2
2.016 g H2
Next determine the limiting factor
2,450 mol CO 2 mol H2 = 4,900 mol H2
1 mol CO
4270 mol H2 < 4,900 mol H2
Hydrogen is the limiting factor.
Next determine the maximum amount of methanol that can be produced in this reaction.
4270 mol H2 1 mol CH3OH = 2140 mol CH3OH
2 mol H2
Convert moles to grams.
2140 mol CH3OH 32.04 g CH3OH= 68600 g CH3OH
1 mol CH3OH
Percent Yield
35700 g CH3OH x 100 % = 52.0%
68600 g CH3OH
FormulasFormulasmolecular formula = (empirical formula)molecular formula = (empirical formula)nn
[[nn = integer] = integer]
molecular formula = Cmolecular formula = C66HH66 = (CH) = (CH)66
empirical formula = CHempirical formula = CH
Empirical formula: the lowest whole number ratio of atoms in a compound.
Molecular formula: the true number of atoms of each element in the
Limiting ReactantLimiting Reactant The The limiting reactantlimiting reactant is the is the
reactant reactant
that is that is consumed firstconsumed first,, limiting limiting the amounts of products the amounts of products formed.formed.
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9-15
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9-17
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9-19
Write and balance the equation for the reaction.
Convert known masses of reactants to moles.
Determine which reactant is limiting.
Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product.
Convert moles to mass (if required).
Write the balanced equation.
N2 + 3H2 2NH3
Convert known masses to moles.
25000 g N2 1 mol N2 = 892 mol N2
28.02 g N2
5000 g H2 1 mol H2 = 2,480 mol H2
2.016 g H2
Determine the Limiting reactant.
892 mol N2 3 mol H2 = 2676 mol H2
1 mol N2
2480 mol H2 < 2676 mol H2
Hydrogen is the limiting reactant. To get all of the Nitrogen to react you would need 2676 mol of hydrogen gas.
Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product.
2480 mol H2 2 mol NH3 = 1653 mol NH3
3 mol H2
Convert moles to mass.
Calculate the molar mass of NH3
N 14.01 x 1 = 14.01
H 1.008 X 3 = 3.024 +
17.034
1653 mol NH3 17.03 g NH3 = 28,150 g NH3
1 mol NH3