Question
Question: Assuming that petrol is octane \({{C}_{8}}{{H}_{18}}\) and has a density of \(0.8gm{{L}^{-1}}\), 1.4...
Assuming that petrol is octane C8H18 and has a density of 0.8gmL−1, 1.425 litre of petrol on combustion will consume:
A. 100 mole of oxygen
B. 125 mole of oxygen
C. 150 mole of oxygen
D. 175 mole of oxygen
Solution
When petrol on combustion produces carbon monoxide, and nitrogen oxides that are poisonous gases. We will find the number of moles by the formula:
Number of moles=Molecular massGiven mass
Complete Solution :
- We are being provided with density= 0.8gmL−1,
- And volume of octane is=1.425 litre
- As we know that formula of density is
density=volumemass
- So, by putting values of density and volume in above formula, we get:
0.8=1.425mass
- So, mass of octane=
0.8×1.425×1000=1140g
- Here, we have converted litre into grams.
Now, we will find the molecular mass of octane:
& {{C}_{8}}{{H}_{18}} \\\ & = 8\times \left( 12 \right)+18\times \left( 1 \right) \\\ & = 96 + 18 \\\ & = 114g \\\ \end{aligned}$$ \- The molecular mass of octane is: 114g \- Now we will calculate the number of moles, $$Number\text{ }of\text{ }moles=\dfrac{Given\text{ }mass}{Molecular\text{ }mass}$$ $$\begin{aligned} & =\dfrac{1140}{114} \\\ & = 10mole \\\ \end{aligned}$$ \- Now, we will see the reaction of combustion of octane: $$2{{C}_{8}}{{H}_{18}}+25{{O}_{2}}\to 16C{{O}_{2}}+18{{H}_{2}}O$$ \- Number of moles of oxygen required for the combustion of one mole octane= 12.5 moles \- Therefore, we can say that the number of moles of oxygen required for the combustion of 10 mole of octane= $\begin{aligned} & 12.5\times 10 \\\ & =125mole \\\ \end{aligned}$ \- Hence, we can conclude that the correct option is (B), that is 1.425 litres of octane will consume about 125 moles of oxygen. **So, the correct answer is “Option B”.** **Note:** \- We should not forget to write units after solving any question. \- Octane is hydrocarbon, which is very harmful, if one swallowed it. It is found to be flammable. It actually forms explosive mixtures when reacted with air.