Download Chapter 4 Quantities of Reactants and Products 4.1 Chemical

Chapter 4 Quantities of Reactants and Products
4.1 Chemical Equations (p. 124)
A chemical equation is a shorthand description of a chemical reaction using symbols and formulas to represent the elements and
compounds involved. It is a symbolic representation of a chemical reaction in terms of chemical formulas.
Reactants -----> Products
or
Reactants = Products
The symbols “------->” or “=”, mean “forms”, “yields”, or “produces.”
The reactants are the starting substances and the products are the substances formed.
The following abbreviations indicate the phases of the reacting participants.
gas =(___); liquid = (___); solid = (___); aqueous = (____)
Combustion of candle wax, C50H102, with molecular oxygen, O2, produces carbon dioxide, CO2, and water, H2O, and is communicated
as follows.
2 C50H102(g) + 51 O2(g) ---> 100 CO2(g) + 102 H2O(g)
Indicate the phases in the following equation.
2Na(__) + 2H2O(__) ----> 2NaOH(___) + H2(__)
____________________ are the numbers placed in front of formulas in a chemical equation to balance the equation, thereby
indicating the combining ratios of the reactants and the products.
Balanced equations obey the law of conservation of mass. That is the total mass before a reaction takes place will equal the total mass
after the reaction is complete. This also holds true for number of atoms involved in the reaction.
For example, the reaction of solid sodium metal with gaseous chlorine is indicated as follows.
2 Na(s) + Cl2(g) ----> 2 NaCl(s)
This balanced reaction can be interpreted on at least three scales or levels:
1. On the smallest scale possible (nanoscale level or atomic level), 2 atoms of sodium react with 1 molecule of chlorine to produce 2
formula units of sodium chloride.
2. On a molar scale, 2 moles of sodium atoms react with 1 mole of chlorine molecules to produce 2 moles of sodium chloride.
3. On a mass scale (practical scale) 45.98 g of sodium react with 70.90 g of chlorine to produce 116.88 g o sodium chloride.
Stoichiometry is the relationship that describes the relation between the masses of reactants and products.
4.2 Patterns of Chemical Reactions (p. 126)
In a combination reaction, two substances combine to form a third substance.
{A + B ----> AB}
In a decomposition reaction, a single compound reacts to give two or more substances.
{AB ----> A + B}
In a displacement reaction (or single replacement reaction), an element reacts with a compound and takes the place of (displaces)
one of the elements in the original compound.
{A + BC -----> AB + C}
An exchange (metathesis reaction or double-replacement reaction), is a reaction between compounds that, when written as a
molecular equation, appears to involve the exchange of parts between the two reactants. Such reactions often occur between ions in
solution that can result in the formation of insoluble substances, weak electrolytes, or nonelectrolytes.
{AX + BY -----> AY + BX}
In combustion reactions a substance reacts with oxygen, usually with the rapid release of heat.
4.3 Balancing Chemical Equations (p. 133)
1. The equation will be understood to proceed from left to right. The reactants are on the left side of the arrow and the products are
on the right.
H2 + Cl2 ----> 2 HCl
2. Equations are balanced by adjusting coefficients in front of formulas, never by changing subscripts within formulas. Remember
that a 1 is understood when a coefficient is not present.
3. It is best to start with an element that appears in only one compound on each side of the arrow.
4. Next balance any element that appears in more than one compound on either the right or left.
5. Balance free elements last. That is, balance any element that appears in elemental form on the right or left.
6. When polyatomic species (NH4+, SO4-2, OH-, etc.) appear on both sides of the arrow in compounds or as ions, balance them as units
rather than individual elements.
7. Leave the coefficients in the lowest whole number ratio.
Balance the following equations.
#1
H2 + O2 ------> H2O
#2
N2 + H2 ------>
#3
Na(s) + H2O(l) ---> NaOH(aq) + H2(g)
#4
H3PO4 + Ca(OH)2 ----> Ca3(PO4)2 + H2O
#5
Ca3(PO4) 2 + SiO2 +
#6
As2S3 + O2 ----> As2O3 + SO2
NH3
C ---> P4 +
CaSiO3 +
CO
Balance the following combustion of hydrocarbon rxns.
#1
CH4 + O2 ---> CO2 + H2O
#2
C2H4 +
O2 --->
CO2 +
H2O
#3
C2H6 +
O2 --->
CO2 +
H2O
#4
C4H8 +
O2 --->
CO2 +
H2O
#5
C5H10 +
#6
C50H102 +
#7
Balance the equation for the combustion of C6H6 .
#8
What is the sum of the coefficients in the balanced equation for the combustion of heptane, C7H14?
O2 --->
O2 --->
4.4 The Mole and Chemical Reactions: The Macro-Nano Connection (p. 136)
A balanced equation can be used to derive molar ratios of the participants in the reaction.
2 C4H10(g) + 13 O2(g) ---> 8 CO2(g) + 10 H2O(g)
Using the equation above, derive the following relationships.
_____mol O2/ _____mol C4H10 or _____mol CO2/ _____mol C4H10 or _____mol H2O/ _____mol O2
Stoichiometry Problem #1
How many moles of oxygen are required to react with 0.50 mol of butane, C4H10? [3.25 mol O2]
Stoichiometry Problem #2
How many moles of water can be formed from the complete reaction of 3 moles of oxygen with an excess of hydrogen? [6 moles]
Stoichiometry Problem #3
How many moles of water can be formed from the reaction of 0.250 moles of phosphoric acid with sufficient calcium hydroxide?
[0.75 moles]
2 H3PO4 + 3 Ca(OH)2 ----> Ca3(PO4)2 + 6 H2O
Stoichiometry Problem #4
How many moles of phosphoric acid are required to react with 15.00 g of calcium hydroxide? [Molar mass of calcium hydroxide is
74.10 g/mol.] [0.1350 mol]
2 H3PO4 + 3 Ca(OH)2 ----> Ca3(PO4)2 + 6 H2O
Stoichiometry Problem #5
What mass of oxygen is needed to combust 1.00 g of C6H6? [3.072g]
Stoichiometry Problem #6
What mass of water could be formed from the complete combustion 35.00 g of C8H18? [49.68 g]
4.5 Reactions with One Reactant in Limited Supply (p. 141)
A balanced equation shows the relative proportions between reactants and between products. This relation is fixed but the actual
amounts of reactants present can vary. This means that sometimes there can be an excess of one or more of the reactants. In other
words the quantity of one reactant will control the amount of products that will be formed.
The following steps can be used to calculate the mass produced of a given substance when one or more of the reactants are present in a
limiting mass.
Step 1. Calculate the amount of a specific product (moles or grams, as needed) that can be formed from each reactant.
Step 2. The reactant that gives the least amount of product is the limiting reactant. The limiting reactant will determine the amount of
product formed in the reaction.)
Limiting Reagent Problem #1
How many moles of H2O will be formed when 4.0 moles of H2 are allowed to react with 1.0 mole of O2? [2.0 mol H2O]
2 H2 + O2 ----> 2 H2O
Limiting Reagent Problem #2
How many moles of BaSO4 will be produced from a mixture of 3.5 moles H2SO4 and 2.5 moles BaCl2? (2.5 moles BaSO4)
H2SO4 + BaCl2 -----> BaSO4 + 2 HCl
Limiting Reagent Problem #3
A mixture of 35.0 g of hydrogen and 270 grams of oxygen react to form water. How many grams of water will form? (304 g H2O)
Limiting Reagent Problem #4
How many grams of silicon will form when 15.0 grams of boron are reacted with 50.0 grams of silicon dioxide according to the
following equation? (23.4 g Si)
4 B + 3 SiO2 -----> 3 Si + 2 B2O3
Limiting Reactant Demonstration
The reaction of Vinegar and Sodium Bicarbonate
7 bottles contain 0.25 mol of acetic acid.
Balloons are attached containing 3g, 6g, 12g, 18g, 24g, 30g, and 36g of sodium bicarbonate (FW = 84.01).
#1 What is the limiting reagent in each bottle?
#2 How much carbon dioxide will be produced in each bottle?
HC2H3O2(l) + NaHCO3(s) ------> NaC2H3O2(aq) + H2O(l) + CO2(g)
Bottle
1
2
3
4
5
6
7
Moles HC2H3O2
0.25 mol
0.25 mol
0.25 mol
0.25 mol
0.25 mol
0.25 mol
0.25 mol
Moles NaHCO3
3.00 g = 0.0357 mol
6.00 g = 0.0714 mol
12.0 g = 0.143 mol
18.0 g = 0.214 mol
24.0 g = 0.286 mol
30.0 g = 0.357 mol
36.0 g = 0.429 mol
Limiting Reagent
Moles CO2 (g)
4.6 Evaluating the Success of a Synthesis: Percent Yield (p. 146)
Theoretical yield is the calculated maximum amount of product that can be obtained from a given amount of reactant, according to
the chemical equation. The actual yield is the measured amount of product that is finally obtained. The actual yield is often less than
the theoretical yield. (And certainly is never more!) The percent yield is the ratio of the actual yield to the theoretical yield
multiplied by 100.
Percent yield =
actual yield x 100
theoretical yield
Percent Yield Problem #1
If the theoretical yield calculated for a reaction is 14.8 g, and the amount of product obtained is 9.25 g, what is the percent yield?
[62.5%]
Percent Yield Problem #2
Carbon tetrachloride, CCl4, was prepared by reacting excess carbon disulfide, CS2, with 100. g of chlorine. Calculate the percent yield
if 65.0 g of CCl4 was obtained from the reaction. [89.9%]
CS2 + 3 Cl2 ---> CCl4 + S2Cl2
Percent Yield Problem #3
Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the
percent yield if 375.0 g of silver bromide was obtained from the reaction. [91.9%]
MgBr2 + 2 AgNO3 ---> Mg(NO3)2 + 2 AgBr
Percent Yield Problem #4 [Hint: This involves a limiting reagent problem!]
When 100 g of oxalic acid (H2C2O4) and 100 g of methanol (CH3OH) were allowed to react, 38 g of methyl oxalate (C4H6O4) were
isolated. What is the percent yield? [29%]
H2C2O4 +
oxalic acid
2 CH3OH
methanol
<--->
C4H6O4 + 2 H2O
methyl oxalate
4.7 Percent Composition and Empirical Formulas (p. 150)
In a combustion analysis of a compound containing carbon and hydrogen, the compound reacts with oxygen and all of the carbon in
the compound is converted to carbon dioxide and the hydrogen in the compound is converted to water.
2 C4H10(g) + 13 O2(g) ---> 8 CO2(g) + 10 H2O(g)
The law of conservation of mass and the concept of following the mass of elements from one single compound to several compounds
provide a basis for determining percent composition, empirical formula and molecular formula.
Combustion Analysis Problem #1
Benzene is a liquid compound composed of carbon and hydrogen. A sample of benzene weighing 342 mg is burned in excess oxygen
and forms 1156 mg of carbon dioxide. What is the percent composition of benzene? [92.25%C & 7.75% H]
Combustion Analysis Problem #2
n-Butyl phthalate is used as an insect repellant and is composed of carbon, hydrogen, and oxygen. When a 0.3413 g sample was
subjected to combustion analysis, 0.2430 g of water and 0.8633 g of carbon dioxide were produced. In another analysis, the molecular
weight was determined to be 278.38 g/mol. Calculate the empirical formula and the molecular formula. [C8H11O2 & C16H22O4]
Combustion Analysis Problem #3
A 0.1034 g sample of a compound composed of carbon, hydrogen and oxygen is subjected to combustion analysis, producing 0.2351g
CO2 and 0.0962 g H2O. What is the empirical formula for the compound? [C3H6O]
If the molecular weight is 116.18 g/mol, what is the molecular formula? [C6H12O2]