chemical reactions. balancing chemical equations- problem sodium metal reacts with water to produce...

Chemical Chemical ReactionsReactions

Balancing Chemical Balancing Chemical Equations- ProblemEquations- Problem

Sodium metal reacts with water to Sodium metal reacts with water to produce aqueous sodium hydroxide produce aqueous sodium hydroxide and hydrogen.and hydrogen.



Na(s) + HNa(s) + H22O(O(l l ) )



Na(s) + HNa(s) + H22O(O(l l ) ) NaOH(aq) NaOH(aq) + H+ H22(g)(g)


Sodium metal reacts with water to produce Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen.aqueous sodium hydroxide and hydrogen.

Na(s) + HNa(s) + H22O(O(l l ) ) NaOH(aq) + H NaOH(aq) + H22(g)(g)

The equation is not yet balanced. Hydrogens The equation is not yet balanced. Hydrogens come in twos on the left, and three come in twos on the left, and three hydrogens are on the right side of the hydrogens are on the right side of the equation.equation.



Na(s) + HNa(s) + H22O(O(l l ) ) NaOH(aq) NaOH(aq) + H+ H22(g)(g)

Try a “2” in front of the water.Try a “2” in front of the water.


Sodium metal reacts with water to Sodium metal reacts with water to produce aqueous sodium hydroxide and produce aqueous sodium hydroxide and hydrogen.hydrogen.

Na(s) + 2 HNa(s) + 2 H22O(O(l l ) ) NaOH(aq) + NaOH(aq) + HH22(g)(g)

We now have two O atoms on the left, so We now have two O atoms on the left, so we need to put a 2 before NaOH.we need to put a 2 before NaOH.



Na(s) + 2 HNa(s) + 2 H22O(O(l l ) ) 2NaOH(aq) + 2NaOH(aq) + HH22(g)(g)

The two sodium atoms on the right require The two sodium atoms on the right require that we put a 2 in front of Na on the left.that we put a 2 in front of Na on the left.


Sodium metal reacts with water to produce Sodium metal reacts with water to produce aqueous sodium hydroxide and hydrogen.aqueous sodium hydroxide and hydrogen.

2 Na(s) + 2 H2 Na(s) + 2 H22O(O(l l ) ) 2NaOH(aq) + 2NaOH(aq) + HH22(g)(g)

The two sodium atoms on the right require The two sodium atoms on the right require that we put a 2 in front of Na on the left. that we put a 2 in front of Na on the left. The equation is now balanced.The equation is now balanced.




Left SideLeft Side Right SideRight SideNa- 2Na- 2 Na- 2Na- 2H- 4H- 4 H- 4H- 4O- 2O- 2 O- 2O- 2

Chemical EquationsChemical Equations


The balanced chemical equation The balanced chemical equation can be interpreted in a variety of ways. can be interpreted in a variety of ways.

It could say that 2 atoms of sodium It could say that 2 atoms of sodium react with 2 molecules of water to react with 2 molecules of water to produce 2 molecules of sodium produce 2 molecules of sodium hydroxide and a molecule of hydrogen.hydroxide and a molecule of hydrogen.



The balanced chemical equation The balanced chemical equation can be interpreted in a variety of ways. can be interpreted in a variety of ways.

It could say that 200 atoms of It could say that 200 atoms of sodium react with 200 molecules of sodium react with 200 molecules of water to produce 200 molecules of water to produce 200 molecules of sodium hydroxide and 100 molecules of sodium hydroxide and 100 molecules of hydrogen.hydrogen.


2 Na(s) + 2 H2 Na(s) + 2 H22O(O(l l ) ) 2NaOH(aq) + H2NaOH(aq) + H22(g)(g)The balanced chemical equation can be The balanced chemical equation can be interpreted in a variety of ways. interpreted in a variety of ways.

It is usually interpreted as 2 moles of It is usually interpreted as 2 moles of sodium will react with 2 moles of water to sodium will react with 2 moles of water to produce 2 moles of sodium hydroxide and produce 2 moles of sodium hydroxide and 1 mole of hydrogen.1 mole of hydrogen.

The balanced equation tells us nothing The balanced equation tells us nothing about the about the massesmasses of reactants or of reactants or products.products.

Types of Chemical Types of Chemical ReactionsReactions

A A combination reactioncombination reaction is a reaction is a reaction with two or more reactants, and a with two or more reactants, and a single product.single product.

2 CO2 CO ((gg) + O) + O22((gg) ) 2 CO 2 CO22((gg) )


A A decomposition reactiondecomposition reaction is a is a reaction has one reactant, and two reaction has one reactant, and two or more products.or more products.

CaCOCaCO33((ss) ) CaO( CaO(ss) + CO) + CO22((gg) )


In a In a combustion reactioncombustion reaction, a substance , a substance burns in the presence of oxygen. If the burns in the presence of oxygen. If the reactant is a compound containing on reactant is a compound containing on carbon, hydrogen and oxygen, the carbon, hydrogen and oxygen, the products are water and carbon dioxide.products are water and carbon dioxide.

CHCH44((gg) +2 O) +2 O22((gg) ) 2 H 2 H22O (O (ll) + CO) + CO22((gg) )

Chemical CompositionChemical Composition

Usually, the compound is Usually, the compound is combusted in the presence of oxygen. combusted in the presence of oxygen. Any carbon in the compound is Any carbon in the compound is collected as carbon dioxide (COcollected as carbon dioxide (CO22), and ), and any hydrogen is collected as water any hydrogen is collected as water (H(H22O).O).

Determining Empirical Determining Empirical FormulasFormulas

If given combustion dataIf given combustion data::

The ultimate goal is to get the The ultimate goal is to get the simplest whole number ratio of the simplest whole number ratio of the elements in the compound. Usually elements in the compound. Usually the compound contains carbon, the compound contains carbon, hydrogen and perhaps oxygen or hydrogen and perhaps oxygen or nitrogen.nitrogen.

1. Use the information about CO1. Use the information about CO22 to to determine the moles and mass of determine the moles and mass of carbon in the compound.carbon in the compound.

Determining Empirical Determining Empirical FormulasFormulas

If given combustion dataIf given combustion data::

2. Use the information about H2. Use the information about H22O to O to determine the moles and mass of determine the moles and mass of

hydrogen hydrogen in the compound.in the compound.

3. The mass and moles of oxygen (or a 3. The mass and moles of oxygen (or a third element) can be obtained by third element) can be obtained by difference.difference.

4. Once moles of each element is 4. Once moles of each element is obtained, find the relative number of obtained, find the relative number of moles and empirical formula (as with moles and empirical formula (as with % composition).% composition).

Formulas Formulas from from Combustion Combustion DataData

This method assumes the compound contains only C and H.

Empirical Formula using Empirical Formula using Combustion Data- ProblemCombustion Data- Problem

A compound, which contains C, H A compound, which contains C, H and O, is analyzed by combustion. If and O, is analyzed by combustion. If 10.68 mg of the compound produces 10.68 mg of the compound produces 16.01 mg of carbon dioxide and 4.37 16.01 mg of carbon dioxide and 4.37 mg of water, determine the mg of water, determine the empirical formula of the compound.empirical formula of the compound.

If the compound has a molar mass If the compound has a molar mass of 176.1 g/mol, determine the of 176.1 g/mol, determine the molecular formula of the compound.molecular formula of the compound.

StoichiometryStoichiometry

Stoichiometry is a Greek word that Stoichiometry is a Greek word that means using chemical reactions to means using chemical reactions to calculate the amount of reactants calculate the amount of reactants needed and the amount of products needed and the amount of products formed.formed.

Amounts are typically calculated in Amounts are typically calculated in grams (or kg), but there are other grams (or kg), but there are other ways to specify the quantities of ways to specify the quantities of matter involved in a reaction.matter involved in a reaction.

StoichiometryStoichiometry

A balanced chemical equation or A balanced chemical equation or reaction is needed before any reaction is needed before any calculations can be made.calculations can be made.

The formulas of all reactants The formulas of all reactants and products are written before and products are written before attempting to balance the equation.attempting to balance the equation.

Stoichiometry ProblemStoichiometry Problem

Sodium metal reacts with water to Sodium metal reacts with water to produce aqueous sodium hydroxide produce aqueous sodium hydroxide and hydrogen. How many grams of and hydrogen. How many grams of sodium are needed to produce 50.0g sodium are needed to produce 50.0g of hydrogen?of hydrogen?

A balanced chemical equation is A balanced chemical equation is needed before any calculations can needed before any calculations can be made.be made.



Na(s) + HNa(s) + H22O(O(l l ) ) NaOH(aq) + NaOH(aq) + HH22(g)(g)



2 Na(s) +2 H2 Na(s) +2 H22O(O(l l ) ) 2 NaOH(aq) 2 NaOH(aq) + H+ H22(g)(g)



How many grams of sodium are How many grams of sodium are needed to produce 50.0g of hydrogen?needed to produce 50.0g of hydrogen?


? grams? grams 50.0 g 50.0 g



? grams? grams 50.0 g50.0 g

Although the question doesn’t state it, Although the question doesn’t state it, you can assume enough water is present you can assume enough water is present for complete reaction.for complete reaction.

We can map out the problem:We can map out the problem:

g Hg H22 moles H moles H22 moles Na moles Na grams Na grams Na



? grams? grams 50.0 g 50.0 g We can map out the problem:We can map out the problem:


We use the molar mass of HWe use the molar mass of H22 to go from to go from grams of Hgrams of H22 to moles of H to moles of H22..


2 Na(s) + 2 H2 Na(s) + 2 H22O(O(l l ) ) 2NaOH(aq) + H2NaOH(aq) + H22(g)(g)



molar mass of Hmolar mass of H22





molar mass of Hmolar mass of H22

We use the coefficients from the balanced We use the coefficients from the balanced equation to go from moles of Hequation to go from moles of H22 to moles of to moles of Na.Na.



? grams? grams 50.0 g50.0 g We can map out the problem:We can map out the problem:

g Hg H22 moles H moles H22 moles Na moles Na grams grams NaNa

molar mass of Hmolar mass of H2 2 coefficientscoefficients





molar mass of Hmolar mass of H2 2 coefficientscoefficients We use the molar mass of Na to go from We use the molar mass of Na to go from

moles of Na to grams of Na.moles of Na to grams of Na.




g Hg H22 moles H moles H22 moles Na moles Na grams grams NaNa

molar mass of Hmolar mass of H2 2 coefficients molar mass coefficients molar mass of Na of Na



? grams? grams 50.0 g50.0 g


molar mass of Hmolar mass of H2 2 coefficients molar mass coefficients molar mass of Na of Na

(50.0 g H(50.0 g H22) ) (1 mol H(1 mol H22)) (2 moles Na)(2 moles Na) ( 22.99 g Na)( 22.99 g Na) (2.02 g H(2.02 g H22) (1 mol H) (1 mol H22) (1 mol Na)) (1 mol Na)



(50.0 g H(50.0 g H22) ) (1 mol H(1 mol H22)) (2 moles Na)(2 moles Na) ( 22.99 g ( 22.99 g

Na)Na) = = (2.02 g H(2.02 g H22) (1 mol H) (1 mol H22) (1 mol Na)) (1 mol Na)

= 1,138 grams Na = 1.14 x 10= 1,138 grams Na = 1.14 x 103 3 g Na = g Na = 1.14 kg Na1.14 kg Na

Limiting Reagent Limiting Reagent ProblemsProblems

Sometimes you are given quantities of Sometimes you are given quantities of more than onemore than one reactant, and asked to reactant, and asked to calculate the amount of product formed. calculate the amount of product formed. The quantities of reactants might be The quantities of reactants might be such that both react completely, or one such that both react completely, or one might react completely, and the other(s) might react completely, and the other(s) might be in excess. These are called might be in excess. These are called limiting reagentlimiting reagent problems, since the problems, since the quantity of one of the reacts will limit the quantity of one of the reacts will limit the amount of product that can be formed.amount of product that can be formed.

Limiting Reagent - Limiting Reagent - ProblemProblem

Aluminum and bromine react to form Aluminum and bromine react to form aluminum bromide. If 500. g of aluminum bromide. If 500. g of bromine are reacted with 50.0 g of bromine are reacted with 50.0 g of aluminum, what is the theoretical aluminum, what is the theoretical yield of aluminum bromide?yield of aluminum bromide?1. First write the formulas for reactants 1. First write the formulas for reactants

and products.and products.

Al + BrAl + Br22 AlBr AlBr33


Aluminum and bromine react to form Aluminum and bromine react to form aluminum bromide. If 500. g of aluminum bromide. If 500. g of bromine are reacted with 50.0 g of bromine are reacted with 50.0 g of aluminum, what is the theoretical aluminum, what is the theoretical yield of aluminum bromide?yield of aluminum bromide?2. Now balance the equation by adding 2. Now balance the equation by adding

coefficients.coefficients.

2 Al +3 Br2 Al +3 Br22 2 AlBr2 AlBr33


Aluminum and bromine react to form Aluminum and bromine react to form aluminum bromide. If 500. g of aluminum bromide. If 500. g of bromine are reacted with 50.0 g of bromine are reacted with 50.0 g of aluminum, what is the theoretical aluminum, what is the theoretical yield of aluminum bromide?yield of aluminum bromide?


The The theoretical yieldtheoretical yield is the maximum is the maximum amount of product that can be formed, amount of product that can be formed, given the amount of reactants. It is given the amount of reactants. It is usually expressed in grams.usually expressed in grams.



Given: 50.0gGiven: 50.0g 500.g500.g ? ? gramsgrams

There are several ways to solve this problem. There are several ways to solve this problem. One method is to solve the problem twice. One method is to solve the problem twice. Once, assuming that all of the aluminum Once, assuming that all of the aluminum reacts, the other assuming that all of the reacts, the other assuming that all of the bromine reacts. The correct answer is bromine reacts. The correct answer is whichever assumption provides the whichever assumption provides the smallest smallest amountamount of product. of product.



Given: 50.0gGiven: 50.0g 500.g500.g ? ? gramsgrams

The problem can be mapped:The problem can be mapped:

Al: grams Al Al: grams Al moles Al moles Al moles AlBr moles AlBr33 g AlBrg AlBr33

molar mass Al coefficients molar molar mass Al coefficients molar mass mass AlBrAlBr33

Limiting Reagent - Limiting Reagent - ProblemProblem2 Al +3 Br2 Al +3 Br22 2 AlBr2 AlBr33

Given:Given: 50.0g50.0g 500.g ? grams 500.g ? grams

The problem can be mapped:The problem can be mapped:

Al: grams Al Al: grams Al moles Al moles Al moles AlBr moles AlBr33 g AlBr g AlBr33

molar mass Al coefficients molar mass molar mass Al coefficients molar mass AlBrAlBr33

(50.0 g Al) (50.0 g Al) ( 1 mol Al )( 1 mol Al ) ( 2 mol AlBr( 2 mol AlBr33)) (266.7 g (266.7 g AlBrAlBr33))

(26.98 g Al) ( 2 mol Al) (mol (26.98 g Al) ( 2 mol Al) (mol AlBrAlBr33) )



Given:Given: 50.0g 500.g50.0g 500.g ? ? gramsgrams

Al: Al: (50.0 g Al) (50.0 g Al) ( 1 mol Al )( 1 mol Al ) ( 2 mol AlBr( 2 mol AlBr33)) (266.7 g AlBr(266.7 g AlBr33))

(26.98 g Al)(26.98 g Al) ( 2 mol ( 2 mol Al) (mol AlBrAl) (mol AlBr33) )

= 494. g AlBr= 494. g AlBr33 (if all of the Al reacts) (if all of the Al reacts)



Given:Given: 50.0g 500.g50.0g 500.g ? grams ? grams

The calculation is repeated for BrThe calculation is repeated for Br22..

g Brg Br22 moles Br moles Br22 moles AlBr moles AlBr33 g g AlBrAlBr33

molar mass Brmolar mass Br22 coefficients molar coefficients molar mass AlBrmass AlBr33



Given:Given: 50.0g 500.g50.0g 500.g ? grams ? grams

The calculation is repeated for BrThe calculation is repeated for Br22..

g Brg Br22 moles Br moles Br22 moles AlBr moles AlBr33 g AlBr g AlBr33

molar mass Brmolar mass Br22 coefficients molar mass AlBr coefficients molar mass AlBr33

(500. g Br(500. g Br22) ) (1 mol Br(1 mol Br22) ) (2 moles AlBr(2 moles AlBr33)) (266.7 g (266.7 g AlBrAlBr33)) (159.8 g Br(159.8 g Br22) (3 moles Br) (3 moles Br22) (1 mol ) (1 mol AlBrAlBr33))

= 556. grams of AlBr= 556. grams of AlBr33



Given: 50.0gGiven: 50.0g 500.g500.g ? grams ? grams Summary:Summary:

We have enough Al to produce 494. g We have enough Al to produce 494. g AlBrAlBr33

We have enough BrWe have enough Br22 to produce 556. grams of to produce 556. grams of AlBrAlBr33

The theoretical yield is 494. grams of AlBrThe theoretical yield is 494. grams of AlBr33..



Given: 50.0g 500.g 494. gGiven: 50.0g 500.g 494. gSummary:Summary:

All of the Al reacts, so Al is limiting.All of the Al reacts, so Al is limiting. Bromine is in excess.Bromine is in excess. Additional questions:Additional questions:

1. How much bromine is left over?1. How much bromine is left over?2. If 418 grams of AlBr2. If 418 grams of AlBr33 is obtained, is obtained,

what is the % yield?what is the % yield?



Given: Given: 50.0g 500.g 494. g 50.0g 500.g 494. g

1. How much bromine is left over?1. How much bromine is left over?

Since all 50.0 g of the Al reacts, the Since all 50.0 g of the Al reacts, the product must contain 494.g -50.g = product must contain 494.g -50.g = 444. g of bromine. Therefore, 500.g- 444. g of bromine. Therefore, 500.g- 444.g = 56. g of Br444.g = 56. g of Br22 are left over. are left over.



Given: Given: 50.0g 500.g 494. g 50.0g 500.g 494. g

2.2. If 418 grams of AlBrIf 418 grams of AlBr33 is obtained, what is is obtained, what is the % yield?the % yield?The The percent yieldpercent yield is is (actual yield) (actual yield) (100%) (100%)

(theoretical yield)(theoretical yield)

% yield = % yield = (418 g)(418 g) (100%) = 84.6 % (100%) = 84.6 % (494g)(494g)

Trends in Reactivity of Trends in Reactivity of Main Group ElementsMain Group Elements

In using an element’s position on In using an element’s position on the periodic table to predict reactivity, the periodic table to predict reactivity, there are two underlying factors to there are two underlying factors to consider.consider. The smallest member of a family or group The smallest member of a family or group

often exhibits different behavior than the often exhibits different behavior than the rest of the family due to its small size.rest of the family due to its small size.

Elements on a diagonal often exhibit Elements on a diagonal often exhibit similar behavior even though they are in similar behavior even though they are in different groups.different groups.

Trends in Reactivity of Trends in Reactivity of Main Group ElementsMain Group Elements

For example, both Li and Mg For example, both Li and Mg react with nitrogen to form nitrides react with nitrogen to form nitrides (Li(Li33N and MgN and Mg33NN22), although nitrogen ), although nitrogen is inert with most elements.is inert with most elements.

Metallic CharacterMetallic Character

HydrogenHydrogen

Hydrogen is a Hydrogen is a non metalnon metal. It . It bonds covalently, with the only ionic bonds covalently, with the only ionic form being the hydride ion, Hform being the hydride ion, H––. In . In this way it is similar to the halogens. this way it is similar to the halogens. And is sometimes placed in both And is sometimes placed in both group IA and group VIIA on periodic group IA and group VIIA on periodic tables.tables.

Nancy De Luca

Error in text on page 302. There is no H+ ion found in nature.

Metallic CharacterMetallic Character

Across a period, metallic behavior Across a period, metallic behavior decreases. Non-metals are often crumbly decreases. Non-metals are often crumbly solids, liquids or gases at room temperature. solids, liquids or gases at room temperature.

Metallic Metallic CharacterCharacter

Metallic Metallic behavior increases behavior increases going down a going down a group.group.

Group IA – the Alkali Group IA – the Alkali MetalsMetals

In discussing the chemistry, In discussing the chemistry, preparation and properties of the preparation and properties of the group IA elements, it is important to group IA elements, it is important to remember that hydrogen is remember that hydrogen is notnot a a group IA metal. It’s properties and group IA metal. It’s properties and reactivity would place it within reactivity would place it within group 7A (diatomic non-metals), group 7A (diatomic non-metals), rather than group IA.rather than group IA.

Group 1A MetalsGroup 1A Metals

The group 1A metals are soft shiny The group 1A metals are soft shiny metals with fairly low densities (Li, Na metals with fairly low densities (Li, Na and K are less dense than water) and low and K are less dense than water) and low melting points. Sodium melts at 98melting points. Sodium melts at 98ooC, C, and cesium melts at 29and cesium melts at 29ooC.C.

The softness, low density and low The softness, low density and low melting points are the result of weaker melting points are the result of weaker metallic bonding due to only one valence metallic bonding due to only one valence electron in this group.electron in this group.

Group 1A Metals - Group 1A Metals - ProductionProduction

Due to the high reactivity with Due to the high reactivity with oxygen and water, all of the metals are oxygen and water, all of the metals are found in nature in ionic form (Mfound in nature in ionic form (M1+1+). ).

The pure metal must be produced The pure metal must be produced in an oxygen and water-free in an oxygen and water-free environment. Typically, an electrical environment. Typically, an electrical current is passed through the melted current is passed through the melted chloride salt. The metal and the chloride salt. The metal and the chlorine gas are collected separately.chlorine gas are collected separately.

Reactivity TrendsReactivity Trends

The chemical behavior of the The chemical behavior of the group IA metals illustrates periodic group IA metals illustrates periodic trends. As the valence electron trends. As the valence electron occupies a higher quantum level, it occupies a higher quantum level, it experiences less nuclear attraction, experiences less nuclear attraction, and is more easily removed.and is more easily removed.

Group 1A Metals + WaterGroup 1A Metals + Water

The reaction with water forms The reaction with water forms hydrogen gas and the aqueous metal hydrogen gas and the aqueous metal hydroxide. The reaction is so hydroxide. The reaction is so vigorous, that the hydrogen may vigorous, that the hydrogen may ignite.ignite.

2 M(2 M(ss) + 2 H) + 2 H22O(O(ll) ) H H22((gg) + 2 ) + 2 MOH(MOH(aqaq))

Metallic CharacterMetallic CharacterThe group IA metals react with water to The group IA metals react with water to produce hydrogen and the metal hydroxide.produce hydrogen and the metal hydroxide.

Metallic behavior increases going down a group.

chemical reactions. balancing chemical equations- problem sodium metal reacts with water to produce...

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