Chemical Reactions and Chemical Quantities PDF

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Chemical Reactions and Chemical Quantities (Ch 7) Physical changes in matter  Physical change alters the state or appearance of matter, but not its composition o The atoms/molecules that compose a substance do not change their identity  When water boils, it changes from a liquid to a gas state, a physical change Chemical changes in matter  A chemical change alters the composition of matter o Atoms rearrange, transforming the original substances into new substances  A nail rusting is a chemical change o Rust is formed when the iron atom transfers electrons to oxygen atoms to form is the compound iron (III) oxide (Fe2O3) Chemical reactions  A chemical reaction is writtn as a chemical equation o Reactants yield products  Chemical equations provide the folowing information about the reaction: o Molceular or ionic formulas of reactants and products o Physical states of reactants and products o Relative amounts of reactants and prodcts required (in moles or molecules_ Chemical equations: provide all the required informaton  CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (l) o This reaction equation is read as: 1 mole of methane gas reacts with 2 moles of dioxygen gas to form 1 mole of carbon dioxide gas and 2 moles ofliquid water o Knowing moles and molar mas s of the reactants, we can determine weights of reactants required and moles and weight of products made  The atoms in the reactants must be present in the products  The law of conservation of mass  Equations are balanced by balancing atoms The quantities in chemical reactions  Stoichiometry is the study of the numerical relationships between chemical quantities in a chemical atom Balancing chemical equations  Balancing chemical equations required a trial and error approach  General rules to follow for balancing chemical reactions: o Leave elements (atomic and molecular) alone until the end of the process, especially hydrogen or oxygen o Balance any atom that appears in only one reactant and one product. This ratio is fixed, and can't change when you change the coefficients o Next, go to the most complicated compound. Use as a springboard to balance all other compounds Balancing chemical equations  Polyatomic ions that appear unchanged on both sides of the equation can be counted as asingle unit  Adjust atoms from polyatomic elements

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Adjust the atoms from monoatomic elements If you gey stuck, double the most complicated compound and try again Finally, ameks ure that all coefficients are in the lowest possible ratio Balancing chemical equations  Write the balanced chemical equation that represents the combustion of propane o Step 1: write the unbalanced equation o C3H8 (g) + O2 (g) yields (CO2 (g) + H2O (l) o Do an accounting of types and numbers of individual atoms (REQUIRED) o 3-C-1 o 8-H-2 o 2-O-3 o Step 2: balance atoms in the most complicated compound: REQUIRED ACCOUNTING OF NUMBERS  C3H8 (h) ? O2 (g) yields 3CO2 (g) + 4H2O (l)  C3H8 (g) 5 O2 (g) yields (3CO2) (g) + 4H2O (l) o Step 4 Balancing with complex molecules  Write a balanced equation for the metabolism of butanoic acid. Assume the o Do an accounting of types and numbers of individual atoms Balancing ionic equations with polyatomic ions  Write a balanced chemical equation for the aqueous reaction of barium hydroxide and perchloric acid to produce water and barium perchlorate  Strategy: the reactants are Ba(OH)2 and HClO4, and the products are Ba(ClO4)2 and H2O. The reaction is aqueous, all species except H2O will be labeled (aq) in the equation. Being a liquid, H2O will be labeled (L). Write the unbalanced equation  Soltion: Ba(OH)2 (aq) + HClO4 (Aq) yields Ba (ClO4)2 (aq) + H2O (l)  Account o 1 - Ba - 1 o 2-O-1 o 3-H-2 o 1 - ClO4 - 2  One Na appears in one compound on the left and right. They are balanced.  ClO4- appears on both sides of the equation,  We now have a balanced Ba and Cl4-, do another accounting: o Ba(OH)2 (aq) + 2HClO4 (aq) yields Ba (ClO4)2 (aq) + H2O (l) o 1 - Ba - 1 o 2-O-1 o 4-H-2 o 2 - ClO4 - 2  There is twice as much H and I on the elft as on the right. Placing a 2 in front of H2O (l) balances both the //O and H atoms giving us the final balaned equations o Ba(OH)2 (aq) + 2 HclO4 (aq) yields Ba (ClO4 Reaction stoichiometry: what is it about?  The coefficients in a chemical reaction specify the relative amounts in moles of each of the substances involved in the reaction  2 C8H18 (l) + 25 O2 (g) yields 16 CO2 (g) + 18 H2O (g)  This balanced chemical equation says/:

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2 moles of C8H18 reacts with 25 molecules moles of O2 to form 16 molesulces of CO2 and 18 molecules of H2O Mole to mole conversions  2 C8J18 (l) + 25 O2 (g) yields 16 CO2 (g) + 18 H2O (g)  We can use the stoichiometric ratios from a reaction equation as a conversion factor between the molar amount of one ingredient vs. all other ingredients o Stoichiometric ratio: reactant to product  2 moles c8H18 : 16 moles CO2 How many moles CO2 form is 22.0 mole sof C8H18 are combusted (burned) ? Mole to mole and mass to mass conversions  Stoichiometric ratios and molar mass can be used as conversion factors between the mass (grams) of any ingredient and the mass (grams) of any other ingredient  A is reactant and B is the products… The strategy: o Mass of A goes to (molar mass of A) to moles of A o Moles of A (stoichiometric ratio B to A) goes to moles of B o Moles of B (molar mass of B) goes to mass of B Limiting reactant and theoretical yield  1 crsut _ 5 ox tomato sauce +2 cups cheese yields 1 pizza  Using ratios, how many pizzas can you make with 4 crusts, 10 cups of hceese, and 15 oz tomato sauce?  We have enough crusts to make: 4 crusts x (1 pizza/1crust) = 4 pizzas  We have enough cheese to make: 10 cups cheese x (1 pizza/2 cups cheese) = 5 pizzas  We have enough tomato sauce to make: 15 oz sauce x (1 pizza / 5 oz sauce) = 3 pizzas  Therefore, 3 pizzas is the most pizzas we can make. This is the limiting reagent  The limiting reactant is the reactant tha tmake sthe least amoutn of Determining the limiting reactant  Consider th ereaction between 5 moles of CO and 8 moles of H2 to produce methanol. First, write a balanced equation o CO (g) + 2H2 (g) yields CH3OH (l)  How many moles of H2 are necessary in order for 5 moles CO to react? o Moles of H2 = 5 mol CO x (2 mol H2 / 1 mol CO) = 10 mol H2  10 moles of H2 are required, but 8 moles of H2 are available; H2 = limiting reagent o Moles of CO = 8 mol H2 x (1 mol CO / 2 mol H2) = 4 mol CO  Therefore, given 8 moles of H2, 4 Moles of CO are required. o 4 mol CO x 1 mol CH3OH / 1 mol CO = 4 mol CH3OH Alka seltzer (1,700 g sodium bicarbonate and 1.000g citric acid  Find the limiting reactant for the following reaction: o 3NaHCO3 (aq) + H3C6H5O7 (aq) yields 3CO2 (g) + 3H2O (l) + Na3C6H5O7 (aq) o 1.700g NaHCO3 x (1 mol NaHCO3 / 84.01g NaHCO3) o = 0.02024 mol NaHCO3 o 1.000g H3C6H5O7 x (1mol H3C6H5O7 / 192.12 g) o - 0.005205 mol H3C6H6O7 o 0.02024 mol NaHCO3 x (1 mol NaC8H5O7 / 3 mol NaHCO3) = 0.006745 NaC8H6O7 o 0.005205 mol H3C6H5O7 x (1 mol NaC6H5O7 / 1 mol H3C6H5O7) = 0.005205 o Therefore, citric acid is the limiting reactant  Choose one product and determine how many moles Determine the mass of excess NaHCO3 reactant left over

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3NaHCO3 (aq) + H3C6H5O7 (aq) yields 3CO2 + 3H2O (l) + Na3C6H5O7 (aq) o Calculate the amount of NaHCO3 that will react o 0.005205 mol H3C6H5O7 x (3 mol NaHCO3 / 1 mol H3C6H5O7) = 0.01562 mol NaHCO3  If 0.01562 mole mNaHCO3 reacts, we subtract that from the starting quantity 0.02024 mol 0.01562 = -0.01562 0.004620 moles NaHCO3 are unreacted  0.004620 mol NaHCO3 x (84.01 g NaHCO3 / 1 mol NaHCO3 ) = 0.3881 g NaHCO3 unreacted Determine mass of CO2 produced  0.005205 mol H3C6H5O7 x (3 mol CO2 / 1 mol H3C6H5O7) = 0.01562 mol CO2  Convert to grams  0.01562 mol CO2 x (44.01 g CO2 / 1 mol CO2) = 0.6874 g CO2

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More making pizzas  Now assume that while making the 3 pizzas, one pizza is burnt beyond salvation. Only two pizzas are available to eat, the actual yield o The actual yield in a chemical reaction is the actual amount of product made  Think about actual yield in terms of efficiency

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To determine your efficiency in making pizzas, a percentage value can be calculated In chemical reactions, this is called a percent yield Combustion: a type of chemical reaction  A combustion reaction involves the reaction of a substance with O2 to form one or more oxygen containing compounds  Other products of a combustion reaction: o Water (h2O) and heat (energy)  Example: combustion of methane o CH4 (g) + 2 O2 (g) yields CO2 (g) _ 2 H2O (l) Alkali metal reactions  The alkali metals (group 1A) have ns1 outer electron configurations o They form +1 cations to achieve a noble gas configuration o The reactions of the alkali metals with nonmetals are vigorous  Common reaction for alkali metals (M) is with halogens (X) o 2M + 2MX , 2NA (s) + Cl2 (g) yields 2 NaCl (s)  The alkali metals react vigorously with water to form the dissolved alkali metal ion, the hydrozied ion, the hydrogen gas o 2 M(s_ + 2H2O  The reaction is highly exothermic and can be explosive because the heat from the reaction can ignite the hydrogen gas  Groups 1A elements (ns1, n > 2) o Low IE (ionization energy o Never found in nature in pure elemental state o React with oxygen to form metal oxides Halogen reactions  Group 7 elements that have ns2np5 outer electron configurations o Mostly form 1 anions (F only forms - 1 anion) to achieve the "noble gas configuration" o Most reactive of the nonmetal elements  The halogens (X) tend to react with metals especially with group 1 and 2A metals to form ionic compounds such as metal halides (MNx) 2M + nX2 yields 2MXn o Example: 2Fe (s) + 3 CL2 (g) yields 2 FeCl3  These halogens react with hydrogen to form hydrogen halides o 2 (g) + X2  Group 7A elements o All diatomic, but do not exist in elemental form in nature o Form molecular compounds with nonnmetals...