The reaction equated with Method Ping pong

The reaction equated with Method Ping pong

TOPICS: equalizes Reaction Methods Ping Pong

POSTED BY: NANDASANJAYA FEBRUARY 12, 2016

 

The reaction equated with Method Ping pong

 

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Known ping pong method because the process of equalizing our reactions as if they were playing ping pong, namely filling coefficient of one segment and then move to other segments continuously until all the compounds gain coefficient. Consider the following steps.

Balance the reaction: KMnO4 + MnCl2 + Cl2 → HCl + KCl + H2O

1. Give the elements that exist only on one of the compounds on the left and the right side.

KMnO4 + MnCl2 + Cl2 → HCl + KCl + H2O

2. Start by giving the coefficients on compounds that have marked the element most numerous and most indexed. In this example, we will start from KMnO4 because these compounds have a marked element of the index is the most widely and most (4 to O atoms). KMnO4 given a coefficient 2 (1 is fine).

2 KMnO4 + MnCl2 + Cl2 → HCl + KCl + H2O

3. Balance the atoms marked on the left and right.

a. Give coefficient 2 on MnCl2 and KCl to equalize Mn and K.

2 KMnO4 + HCl → 2 MnCl2 + Cl2 + 2 KCl + H2O

b. Balance the O atom to give the coefficient 8 in H2O

2 KMnO4 + HCl MnCl2 + Cl2 → 2 KCl + 8 + 2 H2O

c. Balance the H atoms to give a coefficient of 16 on HCl.

2 KMnO4 + 16 HCl MnCl2 + Cl2 → 2 KCl + 8 + 2 H2O

d. Balance the Cl atom to give the coefficient 5 in Cl2.

2 KMnO4 + 16 2 MnCl2 HCl → Cl 2 + 5 + 2 KCl + 8 H2O

Often the "ping pong game" is not going smoothly until all the compounds has a coefficient. Sometimes a series of coefficients Award interrupted halfway. Consider the following cases.

Case breaking game 1

Balance the reaction: K2Cr2O7 + SnCl2 + HCl → CrCl3 + SnCl4 + KCl + H2O

1. Give the elements that exist only on one of the compounds on the left and the right side.

K2Cr2O7 + SnCl2 + HCl → CrCl3 + SnCl4 + KCl + H2O

2. Start by giving the coefficients on compounds that have marked the element most numerous and most indexed. In this example, K2Cr2O7 given a coefficient of 1.

1 K2Cr2O7 + SnCl2 + HCl → CrCl3 + SnCl4 + KCl + H2O

3. Balance the atoms marked on the left and right.

a. Give coefficient 2 on CrCl3 and KCl to equalize Cr and K.

1 K2Cr2O7 + SnCl2 + HCl → 2 CrCl3 + SnCl4 + 2 KCl + H2O

b. Balance the O atom to give the coefficient 7 in H2O

1 K2Cr2O7 + SnCl2 + HCl → 2 CrCl3 + SnCl4 + 2 KCl + 7 H2O

c. Balance the H atoms to give a coefficient of 14 on HCl.

1 K2Cr2O7 + SnCl2 + 14 + HCl → 2 CrCl3 SnCl4 + 2 KCl + 7 H2O

4. So far the "game of ping pong" we must stop. Still there are two compounds that have not been given a coefficient, which is SnCl2 and SnCl4. Both should have the same coefficient, for example a.

1 K2Cr2O7 + a SnCl2 + 14 + HCl → 2 CrCl3 a SnCl4 + 2 KCl + 7 H2O

To calculate the value of a, consider the Cl atom. The number of Cl atoms on the left and the right should be the same.

2a + 14 = 2 × 3 + 4a + 2 × 1

2a +14 = 4a + 8

a = 3

Thus the end result of the "game" we are:

1 K2Cr2O7 + 3 SnCl2 + 14 HCl → 2 CrCl3 SnCl4 + 2 + 3 + 7 H2O KCl

Case breaking game 2

Balance the reaction: Cus + HNO3 → Cu (NO3) 2 + S + NO + H2O

1. Give the elements that exist only on one of the compounds on the left and the right side.

CUS + HNO3 → Cu (NO3) 2 + S + NO + H2O

2. Start by giving the coefficients on compounds that have marked the element most numerous and most indexed. In this example, cus given a coefficient of 1.

1 Cus + HNO3 → Cu (NO3) 2 + S + NO + H2O

3. Balance the Cu atoms and S. In this case, Cu (NO3) 2 and S each got koefifen 1.

1 Cus + HNO3 → 1 Cu (NO3) 2 + 1 S + NO + H2O

4. At this point we were interrupted game. To continue this game we will give the coefficient on the variable form other compounds.

1 Cus + a HNO3 → 1 Cu (NO3) 2 + 1 S + b NO + ½ a H2O

H2O ½ a given coefficient for the number of H on the left is a.

a. By equating the number of atoms on the left and the right we will get the following relationship.

N: a = 2 + b

b = a - 2 .............................. .. (i)

O: 3a = 6 + b + ½ a

A = 6 + 2½ b ......................... (Ii)

b. Substituting (i) to the equation (ii).

A = 6 + 2½ (a - 2)

1½ a = 4

a = 8/3

c. Substitute a value equal to (i)

b = 8/3 - 2 = 2/3

5. The full reaction to be as follows.

1 8/3 CUS + HNO3 → 1 Cu (NO3) 2 + 1 S + 2/3 + 4/3 H2O NO

So that the coefficient of reaction does not contain fractions, the above reaction multiplied by three.

3 CUS + 8 HNO 3 → 3 Cu (NO3) 2 + 3 S + 2 NO + 4 H2O

Exercises

Balance the following reactions.

1 (NH4) 2Cr2O7 → Cr2O3 + N2 + H2O

2 Fe2S3 + H2O + O2 → Fe (OH) 3 + S

3 MnO2 + H2SO4 + NaI → MnSO4 + Na2SO4 + H2O + I2 (hint: consider SO4sebagai unity)

4 KMnO4 + H2SO4 + H2C2O4 → MnSO4 + K2SO4 + CO2 + H2O (hint: H2C2O4dapat written as H2 (CO2) 2)

5 K2Cr2O7 + FeSO4 + H2SO4 → CR2 (SO4) 3 + Fe2 (SO4) 3 + K2SO4 + H2O

6 KIO3 + KI + HNO3 → I2 + KNO3 + H2O

7 Al + KOH + As2O3 → KAlO2 + AsH3 + H2O

8 FeS + HNO3 → Fe (NO3) 3 + S + NO + H2O

9 As2O3 + KMnO4 + H2SO4 → H3AsO4 + MnSO4 + K2SO4 + H2O

10 CrI3 + H2O2 + KOH → K2CrO4 KIO4 + H2O +

    

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