   Chapter 10, Problem 34PS

Chapter
Section
Textbook Problem

The hydrocarbon octane (C8H18) bums to give CO2 and water vapor:2 C8H18(g) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)If a 0.048-g sample of octane burns completely in O2, what will be the pressure of water vapor in a 4.75-L flask at 30.0 °C? If the O2 gas needed for complete combustion was contained in a 4.75-L flask at 22 °C, what would its pressure be?

Interpretation Introduction

Interpretation:

The pressure of water vapour and oxygen in the reaction of combustion of octane has to be given. Given that the sample has a mass of 0.048g

Concept Introduction:

According to ideal gas equation:

PV=nRT

Numberofmoles=massmolarmass

Explanation

Given that:

Numberofmolesofoctane=massmolarmass=0.048g114g=4.21×10-4mol

The balanced equation for the reaction is:

2C8H18+25O216CO2+18H2O

From this equation one mole of C8H18 produces nine moles of water vapour.  Therefore,

4.21×10-4molC8H18produces3

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