   Chapter 13, Problem 87QAP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
8 views

# Consider the following reaction for the combustion of octane, C 8 H 18 :msp;  2C 8 H 18 ( l ) + 25O 2 ( g )   16CO 2 ( g ) + 18H 2 O ( l ) at volume of oxygen gas at STP would be needed for the complete combustion of 1 0.0  g of octane?

Interpretation Introduction

Interpretation:

The volume of oxygen gas at STP should be calculated.

Concept Introduction:

According to ideal gas equation:

PV=nRT

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

The value of Universal gas constant can be taken as 0.082 L atm K1 mol1.

Mass of gas can be calculated from number of moles and molar mass as follows:

m=n×M

Here, n is number of moles, m is mass and M is molar mass.

Explanation

Given Information:

The mass of octane is 10 g.

Calculation:

The balanced chemical reaction is as follows:

2C8H18(l)+25O2(g)16CO2(g)+18H2O(l)

From the balanced chemical reaction, 2 mol of octane C8H18 reacts with 25 mol of O2. To calculate the volume of oxygen gas, first calculate the number of moles of both C8H18 and oxygen gas.

n=mM

Molar mass of octane C8H18 is 114.23 g/mol thus,

n=10 g114.23 g/mol=0.0875 mol

Since, 2 mol of octane reacts with 25 mol of O2 thus, 1 mol will react with 25/2=12

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