Balance the following equations :

A. NH3 (g) + O2 (g) → NO (g) + H2O (g)

B. C2H6 (g) + O2 (g) → CO2 (g) + H2O (g)

Answer :

A. NH3 (g) + O2 (g) → NO (g) + H2O (g) ( Unbalanced )

Observe that O atoms are found in all but one compound;balance them last. If we balance the H atoms first, then we will know the total number of N atoms in the equation . Therefore, balance H atoms and N atoms before O atoms.

Balance the H atoms by finding the lowest multiple of 3 and 2; this is 6 , place a 2 as the coefficient of NH3 , and a 3 in front of H2O.

2N

__H3__( g) + O2 (g) → NO (g) + 3

__H2__O (g)

Now that we know that 2NH3 molecules are required , we know the number of N atoms needed -2, However , if we place a 2 in front of the NO, that will leave us with 5 O atoms-an odd number. Thus we cannot balance the O atoms without using a fraction, 2,5. The equation can be balanced by placing 2,5 as the coefficient of O2. Fractions can be avoided by changing the coefficients of NH3 and H2O so that an odd number of O atoms does not result. This is accomplished by multiplying the coefficient by 2, yielding 4 and 6.

4N

__H3__(g) + O2 → NO (g) +

__6H2__O (g)

Now we have 12 H atoms on either side , and we know that we need 4 N atoms. Place 4 as the coefficient in front of NO to balance the N atoms.

4

__NH3__(g) + O2 (g) → 4

__N__O (g) + 6

__H2__O (g)

O atoms are the only ones remaining to be balanced. All of the products have their correct coefficients, thus we need 10 O atoms , 4 O from the 4 NO and 6 O from the 6 H2O , only 2 O atoms are on the left side , so we can multiply by 5 to give 10 O.

4

__NH3__(g) + 5

__O2__(g) → 4

__NO__(g) + 6

__H2O__(g) ( Balanced )

Check to see that there are 4 N atoms , 12 H atoms, and 10 atoms on either side of the equation.

B. C2H6 (g) + O2 (g) → CO2 (g) + H2O (g) (unbalanced )

In this equation it is convenient to start with C atoms, proceed to H atoms, and balance the O atoms last.

Two C atoms are on the left they are balanced by placing 2 in front of CO2.

C2H6 (g) + O2 (g) → 2CO2 (g) + H2O (g)

Six H atoms in C2H6 are balanced by placing a 3 in front of the H2O.

__C2H6__(g) + O2 (g) → 2

__C__O2 (g) + 3

__H__2O (g)

A total of 7 O atoms are found in the products. This time lets balance the equation using fractional coefficients. Since there are 2 O atoms on the left, ask yourself "what number multiplied by 2 gives 7 ?" the answer is 3,5, thus the correct coefficient is 3,5.

C2H6 (g) + 3,5O2 (g) → 2CO2 (g) + 3H2O (g) ( Balanced )

A check reveals that 2 C atoms, 6 H atoms, and 7 O atoms are in both the reactants and products.

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