When propan-1-ol is burnt in excess oxygen the only products formed are carbon dioxide and water.

In the balanced equation for this reaction what is the ratio of CO_2 : H_2O molecules formed?

A. 4 : 7

B. 3 : 8

C. 2 : 7

D. 3 : 4

E. 5 : 12

When propan-1-ol is burnt in excess oxygen the only products formed are carbon dioxide and water.

In the balanced equation for this reaction what is the ratio of CO_2 : H_2O molecules formed?

A. 4 : 7

B. 3 : 8

C. 2 : 7

D. 3 : 4

E. 5 : 12

First, we should write down the balanced reaction of propan-1-ol’s complete combustion:

2C_3H_7OH + 9O_2 \rightarrow 8H_2O + 6CO_2

We can see that the ratio here is 6:8, which we can simplify into 3:4. Therefore our answer is D.

4 Likes

Why is it 2 C3H7OH and 9 O2?

Your unbalanced reaction is:

C3H8O + O2 ----> H2O + CO2

let’s balance carbon and Hidrogen: surely on the right member you must have 3 carbon and 8 hidrogen, so:

C3H8O + O2 ----> 4(H2O) + 3(CO2)

Now we need to balance the oxigen: on the left member there are 3 moles of oxigen, instead on the right there are 10. how can we obtain 10 from 3? let’s use a bit of algebra:

we know that C3H8O have 1 oxigen, and that O2 have 2 oxigen;

we have to write:

1+a2 = 10

so

a2 = 9

and so

a= 9/2

we have now the balanced formula:

C3H8O + **9/2**(O2) ----> 4(H2O) + 3(CO2)

finally you have only to multiply the whole formula by 2 (in order to rewrite 9/2 as a Natural number); so:

2[C3H8O + 9/2(O2)] ----> 2[4(H2O) + 3(CO2)]

hence:

2(C3H8O) + 9(O2) ----> 8(H2O) + 6(CO2)

6 Likes

No actually you are right I made a mistake while trying to balance the equation.

You are right, you don’t have to perfectly balance the reaction to solve this question

1 Like