# Chemical Reactions Unit

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Chemical Reactions Unit
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.

The equation above gives the recipe to make one S’more.
S’Mores 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  S2MmOr3 The equation above gives the recipe to make one S’more.

Table 9.1 The equation for a chemical reaction gives the ingredients needed to make a specific amount of product. Copyright © Houghton Mifflin Company

Calculating Masses of Reactants and Products
Balance the equation. Convert mass to moles. Set up mole ratios. Use mole ratios to calculate moles of desired molecule. Convert moles to grams or to particles, if necessary.

Working a Stoichiometry Problem
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 Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al O2  2Al2O3 6.50 g Al 1 mol Al 2 mol Al2O3 g Al2O3 = ? g Al2O3 26.98 g Al 4 mol Al 1 mol Al2O3 6.50 x 2 x ÷ ÷ 4 = 12.3 g Al2O3

Calculating Percent Composition of a Compound
Like all percent problems: part whole Find the mass of each of the components (the elements), Next, divide by the total mass of the compound; then x 100 x 100 % = percent

Example Calculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S. 29.0 g Ag X 100 = 87.1 % Ag 33.3 g total Total = 100 % 4.30 g S X 100 = 12.9 % S 33.3 g total

Percent Yield Theoretical Yield Actual Yield Percent 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

First write the balanced equation. 2H2 + CO  CH3OH
Suppose 68,500 g of CO is reacted with 8,600 g of H2 to produce CH3OH. Calculate the theoretical yield of methanol. If 35,700 of CH3OH is actually produced, what is the percent yield of methanol? 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 H Mol H = 4270 mol H2 2.016 g H2

Cont. 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 = mol CH3OH 2 mol H2

Cont. Convert moles to grams.
2140 mol CH3OH g CH3OH= g CH3OH 1 mol CH3OH Percent Yield 35700 g CH3OH x 100 % = 52.0% 68600 g CH3OH

Formulas molecular formula = (empirical formula)n [n = integer]
molecular formula = C6H6 = (CH)6 empirical 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 Reactant The limiting reactant is the reactant
that is consumed first, limiting the amounts of products formed.

Container (#1) of N2(g) and H2(g).

Before and after the reaction.

Container (#2) of N2(g) and H2(g).

Before and after the reaction.

Figure 9.2: A map of the procedure used in Example 9.7.

Solving a Limiting Reactant Stoichiometry Problem
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).

Suppose 25. 0 kg of nitrogen gas and 5
Suppose 25.0 kg of nitrogen gas and 5.00 kg of hydrogen gas are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion. Write the balanced equation. N2 + 3H2  2NH3 Convert known masses to moles. 25000 g N2 1 mol N = mol N2 28.02 g N2 5000 g H2 1 mol H2 = 2,480 mol H2 2.016 g H2

Cont. 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.

Cont. 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 = mol NH3 3 mol H2

Cont. Convert moles to mass. Calculate the molar mass of NH3
N  x 1 = 14.01 H  X 3 = 17.034 1653 mol NH g NH3 = 28,150 g NH mol NH3