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.
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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)
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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 
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