Question

the combustion of octane, c₈h₁₈, proceeds according to the reaction shown. 2c₈h₁₈(l) + 25o₂(g) → 16co₂(g) + 18h₂o(l) if 394 mol of octane combusts, what volume of carbon dioxide is produced at 38.0°c and 0.995 atm?

Explanation:

Step1: Find moles of \(CO_2\)

From the balanced equation \(2C_8H_{18}(l)+25O_2(g)\to16CO_2(g)+18H_2O(l)\), the mole ratio of \(C_8H_{18}\) to \(CO_2\) is \(2:16 = 1:8\).
If \(n(C_8H_{18})=394\space mol\), then \(n(CO_2)=394\times8 = 3152\space mol\)

Step2: Use the ideal gas law \(PV = nRT\)

First, convert temperature to Kelvin: \(T=(38.0 + 273.15)K=311.15K\)
\(R = 0.0821\space L\cdot atm/(mol\cdot K)\), \(n = 3152\space mol\), \(P=0.995\space atm\)
Rearrange \(PV=nRT\) to solve for \(V\): \(V=\frac{nRT}{P}\)
Substitute the values: \(V=\frac{3152\times0.0821\times311.15}{0.995}\)

Step3: Calculate the volume

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Answer:

\(80934\space L\)