Question

Fill in the coefficients that will balance the following reaction: a0CH3OH + a1O2 → a2CO2 + a3H2O

Answer 1

To find the number of atoms of each element we multiply coefficient × subscript

For example
`5 Ca_1 Cl_2` contains

5 × 1 = 5 ,Ca atoms and

5 × 2 = 10, Cl atoms

If there is a bracket in the chemical formula

For example
`3Ca_3 (P_1 O_4 )_2` we multiply coefficient × subscript × number outside the bracket.......... to find the number of atoms (Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

So 3 × 3 = 9, Ca atoms

3 × 1 × 2 = 6, P atoms

3 × 4 × 2 = 24, O atoms are present.

Let us balance the equation

`CH_3OH + O_2 > CO_2 + H_2O` (Unbalanced)

Reactant Side # of atoms of each element Product Side

1 C 1

4 H 2

3 O 3

Multiplying CH3OH by 2 we will have 2 - C, 8 - H and 6 - O on the reactant side so the equation changes to

`2CH_3OH + O_2 > CO_2 + H_2O` (Unbalanced)

Reactant Side # of atoms of each element Product Side

2 C 1

8 H 2

4 O 3

By multiplying
`CO_2` by 2 and
`H_2 O` by 4 we will have

`2CH_3OH + O_2 >2 CO_2 +4 H_2O` (Unbalanced)

Reactant Side # of atoms of each element Product Side

2 C 2

8 H 8

4 O 8

To balance O atoms on the left we multiply O_2 by 3

`2CH_3OH + 3O_2 >2 CO_2 +4 H_2O` (balanced)(Answer)

Reactant Side # of atoms of each element Product Side

2 C 2

8 H 8

8 O 8

Balanced!!!

Answer 2

a0=2

a1=3

a2=2

a3=4

Step-by-step explanation:

The balanced equation for the complete combustion of methane is:

2CH₃OH₍l₎ + 3O₂₍g₎ → 2CO₂ + 4H₂O₍l₎

This way, there is an equal number of atoms of each kind on both reactants side and the products side.

Therefore the coefficients represented by the unknowns are:

a0=2

a1=3

a2=2

a3=4

The above is the equation for the complete combustion of methanol.