Chemical equations and reaction stoichiometry

Transcription

1 Chemical equations and reaction stoichiometry

2 Chemical Equations reaction stoichiometry: quantitative relationships among substances as y participate in chemical reactions. O C O O O O C O O O Chemical equations represent a very precise, yet a very versatile, language that describes chemical changes

3 Chemical Equations Chemical reactions always involve changing one or more substances into one or more different substances. In or words, chemical reactions rearrange atoms or ions to form or substances. Chemical equations are used to describe chemical reactions, and y show The substances that react, called reactants; The substances formed, called products The relative amounts of substances involved.

4 Chemical Equations We write reactants to left of an arrow and products to right of arrow. The numbers placed in front of compounds in a chemical equation are called coefficients and represent number of molecules of each reactant or product needed to balance equation. Coefficients indicate amount of each compound or element and can be changed to balance equation; omitted coefficients are assumed to be 1. C 4 1 2O 2 88n CO O reactants products Subscripts indicate number of atoms of each element in compound or element and CANNOT be changed when balancing an equation.

5 Balancing Chemical Equations The combustion (burning) of natural gas is a reaction used to heat buildings and cook food. The principal natural gas is methane, C4. The equation that describes reaction of methane with excess oxygen is: heat C 4 1 2O 2 h CO O 1 molecule 2 molecules 1 molecule 2 molecules The arrow may be read yields. The capital Greek letter delta (Δ) is often used in place of word heat. The equation tells us that methane reacts with oxygen to produce carbon dioxide, CO2, and water. More specifically, for every C4 molecule (or mole) that reacts, two molecules (or moles) of O2 also react, and that one CO2 molecule (or mole) and two 2O molecules (or mole) are formed.

6 Balancing Chemical Equations Following Law of Conservation of Matter, re is no detectable change in quantity of matter during an ordinary chemical reaction. This guiding principle is basis for balancing chemical equations and for performing calculations based on those equations. reactants products O C O O + O C O + O O O Because matter is neir created nor destroyed during a chemical reaction, a balanced chemical equation must always include same number of each kind of atom on both sides of equation. When balancing chemical equations, always write equations with smallest possible whole-number coefficients.

7 Balancing Chemical Equations Dimethyl er, C26O, burns in an excess of oxygen to give carbon dioxide and water. Let s balance equation for this reaction. In unbalanced form, equation is: C 2 6 O 1 O 2 h CO O Carbon appears in only one compound on each side, and same is true for hydrogen. We begin by balancing se elements. The subscripts on carbon and hydrogen atoms in C 2 6 O guide us in assigning coefficients on product side: reactants products C 2 6 O O 2 88n 2CO O atom count: 2C, 6, 3O 2C, 6, 7O Remember to multiply all atoms in a compound, including subscript numbers, by coefficient. Do not change formulas of elements or compounds!

8 Balancing Chemical Equations C 2 6 O O 2 88n 2CO O atom count: 2C, 6, 3O 2C, 6, 7O Now re is a different total number of O atoms on each side: 3 on reactant side and 7 on product side. We now need to balance oxygen count by adding 4 more oxygen atoms to reactant side. This is done by adding 2 more O2 molecules (total of 4 oxygen atoms) to one already present giving a total of 3O2 molecules on reactant side: 4 oxygen atoms 2O 2 C 2 6 O O 2 88n 2CO O C 2 6 O 3O 2 88n 2CO O final atom count: 2C, 6, 7O C 2 6 O 3O 2 88n 2C, 6, 7O 2CO O

9 Balancing Chemical Equations C 2 6 O 3O 2 88n 2CO O When we think we have finished balancing, we should always do a complete check that total number of atoms for each of elements on reactant side matches that found on product side of equation. reactants products + + C 2 6 O 3O 2 2CO O final atom count: 2C, 6, 7O 2C, 6, 7O

10 Balancing Chemical Equations Let s examine a reaction that produces a fractional coefficient. Butane, C410, is used in many lighters. The combustion reaction with O2 is initially written in unbalanced form as: C O 2 h CO O Carbon only appears in one compound on each side and same is true for hydrogen. We begin by balancing se elements. The subscripts on carbon and hydrogen atoms in C410 guide us in assigning coefficients on product side: reactants C 4 10 O 2 88n products 4CO O atom count: 4C, 10, 2O 4C, 10, 13O There are different total numbers of O atoms on each side: 2 on reactant and 13 on product side.

11 C410! O2 88n 4CO2! 52O C410! O2 88n 4CO2! 52O Balancing Chemical Equations reactants atom count: 4C, 10, 2O products 5 " 2 # 10 4C, 10, 13O 5 " 2 # 10 are differentatom totalcount: numbers of O2O atoms on each side: 2 on13o reactant and 4C, 10, 4C, 10, oduct side. We need to balance oxygen count by adding 11 oxygen atoms need to balance by oxygen count1½ by adding atoms to reactant side. This is nt We side. This is done adding Ooxygen 2 molecules to one that is cur re are numbers atoms on on each side:side: 2 on reactant and donedifferent by adding atotal total of 6½ or 6.5of O2O molecules reactant giving a total of 6½ or 6.5 O2 molecules on reactant side: product side. We need to balance oxygen count by adding 11 oxygen atoms tant side. This is done by adding 1½ O2 molecules to one that is cur! 6.5 O2on88n 4CO2side:! 52O e, giving a total of 6½ orc6.5 O2 molecules reactant 410 final atom count: 4C, 10, 13O C410! 6.5 O2 88n 4C, 10, 13O 4CO2! 52O Although this is considered a properly balanced equation, chemists do not use fractional final atom count: 4C, 10, 13O 4C, 10, 13O do not ughcoefficients this is considered properly balanced equation, many chemists in chemical aequations. or e fractional coefficients in chemical equations. To convert fractional coef n integer one simply multiplies equation, entire balanced equationdoby ough this isvalue, considered a properly balanced many chemists notorp r value to convert fractional value into To smallest value. Incoeff thi se fractional coefficients in chemical equations. convertinteger fractional

12 2 giving a total of 6½ or 6.5 O side: 2 molecules Balancing Chemical Equations final atom count: 4C, 10, 13O on reactant4c, 10, 13O 6.5 O2 88n 4CO 52O do not o 10! balanced 2!chemists hough this is consideredca4 properly equation, many use fractional coefficients fractional coe final atom count: in 4C,chemical 10, 13Oequations. To convert 4C, 10, 13O an integer value, one simply multiplies entire balanced equation by convertto fractional coefficient intovalue an integer integer entire balanced gertovalue convert fractional intovalue, multiply smallest value. In th equation proper integer value tobalanced convert equation, fractional value intochemists smallestdo integer ugh thismultiply isbyconsidered a properly many not or would entire equation by a factor of 2 to produce final balance value. In this case, multiply entire equation by a factor of 2 to produce final balanced e with fractional coefficients in chemical equations. To convert fractional coef equation with all integer coefficients: all integer coefficients: n integer value, one simply multiplies entire balanced equation by p r value to convert fractional value into smallest integer value. In thi 2 [ C410! 6.5 O2 88n 4CO2! 52O ] ould multiply entire equation by a factor of 2 to produce final balanced ith all integer coefficients: 2C410! 13 O2 88n 8CO2! 102O final atom count: 2[ 20, 26O C48C, 10! 6.5 O2 88n 20, 26O 4CO8C, 2! 52O ] 88n of 8CO 2C4 Let s generate balanced equation foro aluminum metal 10! 13 2 reaction 2! 10 2O with hy oride to produce aluminum chloride and hydrogen. The unbalanced equation final atom count: 8C, 20, 26O 8C, 20, 26O

13 Problem-Solving Tip for Balancing Chemical Equations There is no one best place to start when balancing a chemical equation, but following suggestions might be helpful: 1. Look for elements that appear in only one place on each side of equation (in only one reactant and in only one product), and balance those elements first. 2. A good starting point is usually to pick a pair of compounds with a common element. Then focus on compound with largest subscript for element in question to see if you can use that subscript as coefficient for or compound. 3. If free, uncombined elements appear on eir side, balance m last.

14 Al 1 Cl AlCl3 1 As it now stands, equation doesh not satisfy Law 2 of Conservation of Matter Balancing Chemical Equations Let s generate balanced equation for reaction of aluminum metal with hydrogen because re are two atoms in molecule and three Cl atoms in one formula unit 2 Let sitgenerate balanced equationchloride for reaction ofsatisfy aluminum metal with hydrogen chloride to As now stands, equation does not Law of Conservation of Matte chloride to produce aluminum and hydrogen. The unbalanced equation is(reacof AlCl (product side), but only one atom and one Cl atom in Cl molecule 3 produce aluminum chloride and hydrogen. The unbalanced equation is ecause re are two atoms in 2 molecule and three Cl atoms in one formula un tant side). Alone 1 Cl AlCl f AlClLet s side),chlorine but onlyby atom and one inof Cl molecule (reac front first balance putting ah coefficient ofcl 3 atom in Cl. 3 (product antasside). it now stands, equation does not satisfy of Conservation Matter because re are As it now stands, equation does notlaw satisfy Law ofof Conservation of Matter two first atoms in 2chlorine and three Cl atoms in one formula unit of AlClof side), but only 3 (product Let s balance byinputting coefficient 3products incl front Cl. reactants because re are twomolecule atoms 2 amolecule andof three atoms in one formula unit one atom and one Cl atom in Cl molecule (reactant side). of AlCl (product side),bybut onlyaone atom and one of ClCl. atom in Cl molecule (reaclet s first3 balance chlorine putting coefficient of 3 in front Al! 3Cl 88n AlCl3! 2 tant side). reactants products count:by Al, 3, 3Cl 3Clof Cl. Let s first balanceatom chlorine putting a coefficient of 3Al, in2, front Al! 3Cl 88n AlCl3! 2 Now re are 3 oncount: left Al, and 2 on right. We Al, can2, add hydrogen only as two reactants products atom 3, 3Cl 3Cl atoms at a time (as ) on product side. So we need to find least common multiple Now re are 3 on 2left and 2 on! 3Cl right. We can add hydrogen only as atoms at a time (as Al 88n AlCl 2two 3 a!total of 3 and 2, which is 6, in order to balance atoms. To get of 6 each 2 ) on product side. So we need to find least common multiple of 3 and 2, whichatoms is 6, in on order to Now re are 3 To onatom left 2 on right. We can add hydrogen side, we 3Cl byofand 26 and 3Cl lone 2side, by we 3. This now gives balance multiply atoms. get acount: total atoms on each multiply 3Cl by 2 and only lone as 2 tw Al, 3, Al, 2, 3Cl toms a time (as 2 ) on product side. So we need to find least common multip by 3.at This now gives f 3 and 2, which is 6, in order to balance atoms. To get!a total of 6 atoms on eac 3Cl 3 2 Now re are 3 on leftaland!2 on 88n right. AlCl We can add hydrogen only as two de, we multiply 3Cl by 2 and gives "2 lone 2 by 3. This now"3 atoms at a time (as 2 ) on product side. So we need to find least common multiple of 3 and 2, which is 6, in orderal to balance 88n atoms.alcl To get! a total! 6Cl 32of 6 atoms on each 3 side, we multiply 3Cl by lone This now Al2 and! 3Cl 88n 2 2 by 3. AlCl 3!gives atom count: Al, 6, "2 6Cl Al, 6, 3Cl"3 Unless ot

15 Balancing Chemical Equations Al 3Cl 88n AlCl atom count: Al 6Cl 88n AlCl Al, 6, 6Cl Al, 6, 3Cl Now Cl is again unbalanced (6Cl on left, 3Cl on right), but we can fix this by putting a coefficient of 2 in front of AlCl3 on product side. Al 6Cl 88n AlCl Al 6Cl 88n 2AlCl atom count: Al, 6, 6Cl 2Al, 6, 6Cl Now all elements except Al are balanced (1Al on left, 2Al on right); we complete balancing by putting a coefficient of 2 in front of Al on reactant side. 2 Al 6Cl 88n 2AlCl Al 6Cl 88n 2AlCl final atom count: 2Al, 6, 6Cl 2Al, 6, 6Cl 2Al 6Cl 88n 2AlCl 3 3 2

16 Balancing Chemical Equations Practice exercise Balance following chemical equations 1. P4 + Cl2 PCl3 2. RbO + SO2 Rb2SO3 + 2O 3. P4O10 + Ca(O)2 Ca3(PO4)2 + 2O 4. Na + O2 Na2O 5. Mg3N2 + 2O N3 + Mg(O)2 6. LiCl + Pb(NO3)2 PbCl2 + LiNO3 7. 2O + KO2 KO + O2 8. 2SO4 + N3 (N4)2SO4 9. Na + O2 Na2O2 10. P4 +O2 P4O Ca(CO3)2 + Na2CO3 CaCO3 + NaCO3 12. N3 +O2 NO + 2O 13. Rb + 2O RbO Fe2O3 + CO Fe + CO2 15. (N4)2Cr2O7 N2 + 2O + Cr2O3 16. Al + Cr2O3 Al2O3 + Cr

17 Calculations Based on Chemical Equations Let s use chemical equations to calculate relative amounts of substances involved in chemical reactions. Consider combustion of methane in oxygen. The balanced chemical equation for that reaction is C 4 + 2O 2 CO O ow many O2 molecules react with 47 C4 molecules according to preceding equation? The balanced equation tells us that one C4 molecule reacts with two O2 molecules and yields one CO2 and two molecules of water. We can hence construct proportion 1 mol C 4 47 mol C 4 = 2 mol O 2 X mol O 2 47 mol C 4 mol O 2 = x 2 mol O 2 = 94 mol O 2 1 mol C 4

18 Limiting Reactants Suppose you wish to make several sandwiches using one slice of cheese and two slices of bread for each. Using Bd = bread Ch = cheese Bd2Ch = sandwich recipe for making a sandwich can be represented like a chemical equation: 2Bd + Ch Bd 2 Ch + If you have 10 slices of bread and 7 slices of cheese, you can make only 5 sandwiches and will have 2 slices of cheese left over. The amount of bread available limits number of sandwiches. 10Bd + 7Ch 5Bd 2 Ch + 2Ch A chemical reaction stops as soon as any reactant is consumed, leaving excess reactants as leftovers. The reactant that is completely consumed in a reaction is called limiting reactant because it limits amount of product formed. The or reactants are sometimes called excess reactants

19 Limiting Reactants Suppose, for example, we have a mixture of 10 mol 2 and 7 mol O2, which react to form water: 2 2(g) + O 2(g) 2 2 O (g) Because 2 mol 2 react with 1 mol O2, number of moles of O2 needed to react with 10 mol 2 is Moles of O 2 = ( ) 1 mol O 10 mol 2 2 = 5 mol O 2 2 mol 2 As 7 mol O2 are available before reaction, 7 mol O2-5 mol O2 = 2 mol O2 are present when all 2 is consumed. In this case 2 is limiting reactant while O 2 is excess reactant. Before 10 reaction and 7 O After reaction

20 Limiting Reactants Summarizing data from example before 2 2(g) + O 2(g) 2 2 O (g) Before reaction 10 mol 7 mol 0 mol Change (reaction): -10 mol -5 mol +10 mol After reaction: 0 mol 2 mol 10 mol limiting reactant excess reactant product The second line in table (reaction) summarizes amounts of reactants consumed (minus signs) and amount of product formed (plus sign). These quantities are restricted by quantity of limiting reactant and depend on coefficients in balanced equation. The mole ratio 2:O2:2O = 10:5:10 conforms to ratio of coefficients in balanced equation, 2:1:2.

21 Sample exercise (Limiting Reactants) A process for converting N2 from air into nitrogen-containing compounds is based on reaction of of N2 and 2 to form ammonia (N3): N2(g) + 3 2(g) 2 N3(g) ow many moles of N3 can be formed from 3.0 mol of N2 and 6.0 mol of 2? The number of moles of 2 needed for complete consumption of 3.0 mol of N is: Moles of 2 = ( ) 3 mol 2 3 mol N2 = 9 mol 2 1 mol 2 Because only 6.0 mol 2 are available, we will run out of 2 before N2 is gone, which tells us that 2 is limiting reactant. Therefore, we use quantity of 2 to calculate quantity N3 produced: Moles of N3 = ( ) 2 mol N3 6 mol 2 = 4 mol N3 3 mol 2 Summarizing data N2(g) + 3 2(g) 2 N3(g) Before reaction 3 mol 6 mol 0 mol Change (reaction): - 2 mol - 6 mol + 4 mol After reaction: 1 mol 0 mol 4 mol

22 Sample exercise (Limiting Reactants) The reaction 22(g) + O2(g) 22O(g) is used to produce electricity in a hydrogen fuel cell. Suppose a fuel cell contains 150 g of 2(g) and 1500 g of O2(g). ow many grams of water can form? From balanced equation, we have stoichiometric relations 2mol 2 : 1molO2 : 2mol2O Using molar mass of each substance, we calculate number of moles of each reactant: m mol x Fw 1 mol 2 Moles of 2 = 150 g 2 ( ) = 75 mol g 2 1 mol O 2 Moles of O 2 = 1500 g O 2 ( ) = 47 mol O g O 2 The balanced equation indicates that reaction requires 2 mol of 2 for every mol of O2. Therefore, for all O2 to completely react, we would need 2 x 47 = 94 mol of 2. Since re are only 75 mol of 2, all of O2 cannot react, so it is excess reactant, and 2 is limiting reactant. 22(g) + O2(g) 22O(g) Before reaction 75 mol 47 mol 0 mol Change (reaction): - 75 mol - 37,5 mol + 75 mol After reaction: 0 mol 9,5 mol 75 mol We use limiting reactant 2 to calculate quantity of water formed, 75 mol 2O. Then we can calculate grams of water formed: Grams of 2 O = 75 mol x 18.0 g/mol = 1400 g 2 O

23 Practice exercises 1. When1.50 mol of Al and 3.00 mol of Cl2 combine in reaction 2 Al(s) + 3 Cl2(g) 2 AlCl3(s) a. Which is limiting reactant? b. ow many moles of AlCl3 are formed? c. ow many moles of excess reactant remain at end of reaction? Answers: (a) Al, (b) 1.50 mol, (c) 0.75 mol Cl g of zinc metal are placed in an aqueous solution containing 2.50 g of silver nitrate, reaction is Zn(s) + 2 AgNO3(aq) 2 Ag(s) + Zn(NO3)2(aq) a. Which is limiting reactant? b. ow many grams of Ag form? c. ow many grams of Zn(NO3)2 form? 3. Sodium hydroxide reacts with carbon dioxide as follows: 2 NaO(s) + CO2(g) Na2CO3(s) + 2O(l) Answers: (a)agno3,(b)1.59g,(c)1.39g,(d)1.52gzn a. Which is limiting reactant when 1.85 mol NaO and 1.00 mol CO2 are allowed to react? b. ow many moles of Na2CO3 can be produced? c. ow many moles of excess reactant remain after completion of reaction?