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Given: 2 C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g) . What volume of oxygen is needed to burn 4 moles of gasoline?

User Momouu
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50 moles of O₂ are needed to burn 4 moles of gasoline.

To find the volume of oxygen needed to burn 4 moles of gasoline, we can use the coefficients in the balanced chemical equation to establish a mole ratio between gasoline (C₈H₁₈) and oxygen (O₂).

The balanced equation is:

\[ 2 \text{C}_8\text{H}_{18}(l) + 25 \text{O}_2(g) \rightarrow 16 \text{CO}_2(g) + 18 \text{H}_2\text{O}(g) \]

From the balanced equation, we can see that the mole ratio between C₈H₁₈ and O₂ is 2:25.

So, for every 2 moles of C₈H₁₈, 25 moles of O₂ are required.

If you have 4 moles of C₈H₁₈, the amount of O₂ needed can be calculated as follows:

\[ \text{Moles of O}_2 = \frac{4 \, \text{moles of C}_8\text{H}_{18} \times 25 \, \text{moles of O}_2}{2 \, \text{moles of C}_8\text{H}_{18}} \]

\[ \text{Moles of O}_2 = 4 \times \frac{25}{2} = 50 \]

Therefore, 50 moles of O₂ are needed to burn 4 moles of gasoline.

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