Balancing Chemical Equations, p. 164

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Presentation transcript:

4.1 – Writing and Balancing Chemical Equations — The Magic of Chemistry

Balancing Chemical Equations, p. 164 Chemical equations need to be balanced in order to maintain the conservation laws

Balancing Chemical Equations, p. 164 Law of conservation of mass In a closed system, the total mass does not change The mass of the products equals the mass of the reactants

Balancing Chemical Equations, p. 164 A balanced chemical equation must ensure the conservation of mass, atoms and electrical charges

Balancing Chemical Equations, p. 164 Coefficients Numbers written before the formula of each reactant and product representing how many particles of each are present Apply to or multiple every element and atom in the particle, not just the first

H2S + PbCl2  PbS + HCl H = 2 H = 1 S = 1 S = 1 Pb = 1 Pb = 1 Cl = 2 Cl = 1

H2S + PbCl2  PbS + HCl H = 2 H = 1 S = 1 S = 1 Pb = 1 Pb = 1 Cl = 2 Cl = 1

H2S + PbCl2  PbS + HCl H = 2 H = 1 S = 1 S = 1 Pb = 1 Pb = 1 Cl = 2 Cl = 1

H2S + PbCl2  PbS + HCl H = 2 H = 1 S = 1 S = 1 Pb = 1 Pb = 1 Cl = 2 Cl = 1

H2S + PbCl2  PbS + HCl H = 2 H = 1 S = 1 S = 1 Pb = 1 Pb = 1 Cl = 2 Cl = 1

H2S + PbCl2  PbS + HCl 2 H = 2 H = 1 S = 1 S = 1 Pb = 1 Pb = 1 Cl = 2 Cl = 1

H2S + PbCl2  PbS + HCl 2 H = 2 H = 2 S = 1 S = 1 Pb = 1 Pb = 1 Cl = 2 Cl = 1

H2S + PbCl2  PbS + HCl 2 H = 2 H = 2 S = 1 S = 1 Pb = 1 Pb = 1 Cl = 2 Cl = 2

H2S + PbCl2  PbS + HCl 2 H = 2 H = 2 S = 1 S = 1 Pb = 1 Pb = 1 Cl = 2 Cl = 2

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 13 H = 5 P = 1 P = 1 O = 5 O = 5 Na = 1 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 13 H = 5 P = 1 P = 1 O = 5 O = 5 Na = 1 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 13 H = 5 P = 1 P = 1 O = 5 O = 5 Na = 1 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 13 H = 5 P = 1 P = 1 O = 5 O = 5 Na = 1 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 13 H = 5 P = 1 P = 1 O = 5 O = 5 Na = 1 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 13 H = 5 P = 1 P = 1 O = 5 O = 5 Na = 1 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 13 H = 5 P = 1 P = 1 O = 5 O = 5 Na = 1 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 13 H = 5 P = 1 P = 1 O = 5 O = 5 Na = 1 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 13 H = 5 P = 1 P = 1 O = 5 O = 5 Na = 3 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 13 H = 5 P = 1 P = 1 O = 7 O = 5 Na = 3 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 15 H = 5 P = 1 P = 1 O = 7 O = 5 Na = 3 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 15 H = 5 P = 1 P = 1 O = 7 O = 5 Na = 3 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 1 H = 15 H = 5 P = 1 P = 1 O = 7 O = 5 Na = 3 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 15 H = 5 P = 1 P = 1 O = 7 O = 5 Na = 3 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 15 H = 11 P = 1 P = 1 O = 7 O = 5 Na = 3 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 15 H = 11 P = 1 P = 1 O = 7 O = 5 Na = 3 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 15 H = 11 P = 1 P = 1 O = 7 O = 5 Na = 3 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 15 H = 11 P = 1 P = 1 O = 7 O = 5 Na = 3 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 15 H = 15 P = 1 P = 1 O = 7 O = 7 Na = 3 Na = 3

(NH4)3PO4 + NaOH  Na3PO4 + NH3 + H2O 15 H = 15 P = 1 P = 1 O = 7 O = 7 Na = 3 Na = 3

C19H17NO3 + O2  CO2 + H2O + N2 C = 19 C = 1 H = 17 H = 2 N = 1 N = 2 O = 5 O = 3

C19H17NO3 + O2  CO2 + H2O + N2 C = 19 C = 1 H = 17 H = 2 N = 1 N = 2 O = 5 O = 3

C19H17NO3 + O2  CO2 + H2O + N2 2 C = 19 C = 1 H = 17 H = 2 N = 1 N = 2 O = 5 O = 3

C19H17NO3 + O2  CO2 + H2O + N2 2 C = 19 C = 1 H = 17 H = 2 N = 2 N = 2 O = 5 O = 3

C19H17NO3 + O2  CO2 + H2O + N2 2 C = 38 C = 1 H = 34 H = 2 N = 2 N = 2 O = 8 O = 3

C19H17NO3 + O2  CO2 + H2O + N2 2 C = 38 C = 1 H = 34 H = 2 N = 2 N = 2 O = 8 O = 3

C19H17NO3 + O2  CO2 + H2O + N2 2 38 C = 38 C = 1 H = 34 H = 2 N = 2 N = 2 O = 8 O = 3

C19H17NO3 + O2  CO2 + H2O + N2 2 38 C = 38 C = 38 H = 34 H = 2 N = 2 N = 2 O = 8 O = 3

C19H17NO3 + O2  CO2 + H2O + N2 2 38 C = 38 C = 38 H = 34 H = 2 N = 2 N = 2 O = 8 O = 77

C19H17NO3 + O2  CO2 + H2O + N2 2 38 C = 38 C = 38 H = 34 H = 2 N = 2 N = 2 O = 8 O = 77

C19H17NO3 + O2  CO2 + H2O + N2 2 38 C = 38 C = 38 H = 34 H = 2 N = 2 N = 2 O = 8 O = 77

C19H17NO3 + O2  CO2 + H2O + N2 2 38 C = 38 C = 38 H = 34 H = 2 N = 2 N = 2 O = 8 O = 77

C19H17NO3 + O2  CO2 + H2O + N2 2 38 C = 38 C = 38 H = 34 H = 2 N = 2 N = 2 O = 8 O = 77

C19H17NO3 + O2  CO2 + H2O + N2 2 38 17 C = 38 C = 38 H = 34 H = 2 N = 2 N = 2 O = 8 O = 77

C19H17NO3 + O2  CO2 + H2O + N2 2 38 17 C = 38 C = 38 H = 34 H = 34 N = 2 N = 2 O = 8 O = 77

C19H17NO3 + O2  CO2 + H2O + N2 2 38 17 C = 38 C = 38 H = 34 H = 34 N = 2 N = 2 O = 8 O = 93

C19H17NO3 + O2  CO2 + H2O + N2 2 38 17 C = 38 C = 38 H = 34 H = 34 N = 2 N = 2 O = 8 O = 93

can only add an even amount C19H17NO3 + O2  CO2 + H2O + N2 2 38 17 can only add an even amount C = 38 C = 38 H = 34 H = 34 N = 2 N = 2 O = 8 O = 93 need an odd number

Balancing Chemical Equations, p. 164 If you need an odd amount of a diatomic element, doubling everything else will produce a need for an even amount

C19H17NO3 + O2  CO2 + H2O + N2 2 38 17 C = 38 C = 38 H = 34 H = 34 N = 2 N = 2 O = 8 O = 93

C19H17NO3 + O2  CO2 + H2O + N2 4 76 34 2 C = 38 C = 38 H = 34 H = 34 N = 2 N = 2 O = 8 O = 93

C19H17NO3 + O2  CO2 + H2O + N2 4 87 76 34 2 C = 76 C = 76 H = 68 H = 68 N = 4 N = 4 O = 14 O = 186

C19H17NO3 + O2  CO2 + H2O + N2 4 87 76 34 2 C = 76 C = 76 H = 68 H = 68 N = 4 N = 4 O = 186 O = 186

C19H17NO3 + O2  CO2 + H2O + N2 4 87 76 34 2 C = 76 C = 76 H = 68 H = 68 N = 4 N = 4 O = 186 O = 186

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 1 S = 3 S = 1 O = 16 O = 7 K = 2 K = 2 I = 2 I = 2 H = 2 H = 3

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 1 S = 3 S = 1 O = 16 O = 7 K = 2 K = 2 I = 2 I = 2 H = 2 H = 3

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 1 S = 3 S = 1 O = 16 O = 7 K = 2 K = 2 I = 2 I = 2 H = 2 H = 3

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 1 O = 16 O = 10 K = 2 K = 2 I = 2 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 1 O = 16 O = 10 K = 2 K = 2 I = 2 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 1 O = 16 O = 10 K = 2 K = 2 I = 2 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 16 O = 18 K = 2 K = 6 I = 2 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 16 O = 18 K = 2 K = 6 I = 2 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 16 O = 18 K = 2 K = 6 I = 2 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 16 O = 18 K = 2 K = 6 I = 2 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 16 O = 18 K = 2 K = 6 I = 2 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 5 2 3 Cr = 2 Cr = 2 S = 3 S = 3 O = 16 O = 18 K = 2 K = 6 I = 2 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 5 2 3 Cr = 2 Cr = 2 S = 3 S = 3 O = 16 O = 18 K = 6 K = 6 I = 6 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 5 2 3 Cr = 2 Cr = 2 S = 3 S = 3 O = 16 O = 18 K = 6 K = 6 I = 6 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 5 3 2 3 Cr = 2 Cr = 2 S = 3 S = 3 O = 16 O = 18 K = 6 K = 6 I = 6 I = 2 H = 2 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 5 3 2 3 Cr = 2 Cr = 2 S = 3 S = 3 O = 18 O = 18 K = 6 K = 6 I = 6 I = 2 H = 6 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 5 3 2 3 Cr = 2 Cr = 2 S = 3 S = 3 O = 18 O = 18 K = 6 K = 6 I = 6 I = 2 H = 6 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 5 3 2 3 3 Cr = 2 Cr = 2 S = 3 S = 3 O = 18 O = 18 K = 6 K = 6 I = 6 I = 2 H = 6 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 5 3 2 3 3 Cr = 2 Cr = 2 S = 3 S = 3 O = 18 O = 18 K = 6 K = 6 I = 6 I = 6 H = 6 H = 6

Cr2(SO4)3 + KI + KIO3 + H2O  Cr(OH)3 + K2SO4 + I2 5 3 2 3 3 Cr = 2 Cr = 2 S = 3 S = 3 O = 18 O = 18 K = 6 K = 6 I = 6 I = 6 H = 6 H = 6

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 1 Mo = 1 Cl = 4 Cl = 4 O = 2 O = 1 Ag = 1 Ag = 2

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 1 Mo = 1 Cl = 4 Cl = 4 O = 2 O = 1 Ag = 1 Ag = 2

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 1 Mo = 1 Cl = 4 Cl = 4 O = 2 O = 1 Ag = 1 Ag = 2

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 1 Mo = 1 Cl = 5 Cl = 4 O = 2 O = 1 Ag = 2 Ag = 2

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 1 Mo = 1 Cl = 5 Cl = 4 O = 2 O = 1 Ag = 2 Ag = 2

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 1 Mo = 1 Cl = 5 Cl = 4 O = 2 O = 1 Ag = 2 Ag = 2

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 1 Mo = 2 Cl = 5 Cl = 8 O = 2 O = 1 Ag = 2 Ag = 2

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 1 Mo = 2 Cl = 5 Cl = 8 O = 2 O = 1 Ag = 2 Ag = 2

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 1 Mo = 2 Cl = 5 Cl = 8 O = 2 O = 1 Ag = 2 Ag = 2

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 8 Cl = 8 O = 2 O = 1 Ag = 2 Ag = 2

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 8 Cl = 8 O = 2 O = 1 Ag = 2 Ag = 2

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 16 Cl = 16 O = 2 O = 2 Ag = 4 Ag = 4

MoCl3 + O2 + AgCl  MoCl4 + Ag2O 16 Cl = 16 O = 2 O = 2 Ag = 4 Ag = 4

Keeping Your Balance — Helpful Hints, p. 165 Write a properly balanced equation for the following chemical reaction: Copper(I) oxide and carbon react to form copper and carbon dioxide 2Cu2O + C  4Cu + CO2