Balancing Chemical Equations Obeying the Law of Conservation of Matter/Mass

BCl

3

+ P

4

+ H

2 

BP + HCl

• • Make a T-chart with reactants on the left & products on the right Tally the number of each atom of each element: R_ P B Cl P H 1 3 4 2 1 1 1 1

• • •

BCl

3

+ P

4

+ H

2 

BP + HCl

Notice on the T-chart a noticeable difference between Reactant & Product and start there I would choose P. Write the coefficient 4 in front of BP so that the P’s balance.

BCl 3 + P 4 + H 2  4BP + HCl Recalculate the T-chart. Note that now R’s Boron has 1, but P’s Boron now has 4. Place a coefficient of 4 in front of BCl 3 4BCl 3 + P 4 + H 2  4BP + HCl

#1] 4BCl 3 + P 4 + H 2  4BP + HCl • Update T-chart. B Cl P H R_ 4 12 4 2 P 4 1 4 1   12 12 Note Cl’s lack of balance. Place a 12 before HCl. 4BCl before H 4BCl 3 3 2.

+ P + P 4 4 + H 2 + 6H 2   4BP + 12HCl Note new imbalance of H now. Fix this by adding 6 4BP + 12HCl DONE!!!

•

C

2

H

2

Cl

4

+ Ca(OH)

2 

C

2

HCl

3

+ CaCl

2

+ H

2

O

Fix T-chart. C H Cl Ca O R: 2 P: 2 4 3 4 5 1 1 2 1 Notice that R’s Cl is 4 & therefore P’s Cl must be a multiple of 4. The Cl in C number of Cl.

2 HCl 3 is an odd #. Begin by adding a coefficient of 2 to create an even C 2 H 2 Cl 4 + Ca(OH) 2  2C 2 HCl 3 Recalculate the T-chart.

+ CaCl 2 + H 2 O

#2] C 2 H 2 Cl 4 + Ca(OH) 2  2C 2 HCl 3 + CaCl 2 + H 2 O C R: P: 4 H 2 4 Cl 4 8 Ca 4 1 O 1 1 2 Right away you can see that to balance C & Cl, we need to write a coefficient of 2 before the 1 st cpd.

2C 2 H 2 Cl 4 + Ca(OH) 2  2C 2 HCl 3 + CaCl 2 + H 2 O C H Cl Ca O R: 4 P: 4 6 4  6 8 8 To fix O, add 2 before H 2 O.

1 1 2 1  2 DONE!

#3] (NH

4

)

2

Cr

2

O

7 

N

2

+ Cr

2

O

3

+ H

2

O

• • Fix T-chart.

R: P: R: P: N H Cr O 2 8 2 7 2 2 2 4 Quick fix H 1 st .

(NH 4 ) 2 Cr 2 O 7  N 2 + Cr 2 O 3 + 4H 2 O N H Cr O 2 2 8 8 2 2 7 7 DONE!

#4] Zn

3

Sb

2

+ H

2

O



Zn(OH)

2

+ SbH

3 Zn R: 3 P: 1 Sb 2 1 H 2 5 O 1 2 You can start with Zn or Sb or both simultaneously.

Zn Zn 3 Sb 2 + H 2 O  3Zn(OH) 2 + 2SbH 3 Sb H O R: 3 2 2 1 P: 3 2 12 6

Zn 3 Sb 2 + H 2 O  3Zn(OH) 2 + 2SbH 3 • • Notice that the ratio of H:O is still 2:1 in the new products. Adjust H 2 O’s coefficient and you’re done. Zn 3 Sb 2 + 6 H 2 O  3 Zn(OH) 2 + 2 SbH 3 • See? This really isn’t so hard after all. Take your time & switch between products’ and reactants’ total # of elements.