Question

Write a balanced chemical equation, including physical state symbols, for the combustion of gaseous butane into gaseous carbon dioxide and gaseous water. 2. Suppose 0.360 kg of butane are burned in air at a pressure of exactly 1 atm and a temperature of 20.0 °C. Calculate the volume of carbon dioxide gas that is produced Be sure your answer has the correct number of significant digits.

Answer 1

Answer:1.

1.Balanced equation

C4H10 + 9 02 ==> 5H20 +4CO2

2. Volume of CO2 =596L

Explanation:

1.Combustion of alkane is the reaction of alkanes with Oxygen. And the general equation for the combustion is;

CxHy +( x+y/4) O2 ==> y/2 02 + xCO2

Where x and y are number of carbon and hydrogen atoms respectively.

For butane (C4H10)

x=4 and y=10

Therefore

C4H10 + 9 02 ==> 5H20 +4CO2

2. Mass of butane = 0.360kg

Molar mass of C4H10 = ( 12×4 + 1×10)

= 48 +10=58g/mol= 0.058kg/mol

Mole = mass/molar mass

Mole = 0.360/0.058= 6.2moles

From the stoichiometric equation

1mole of C4H10 will gives 4moles of CO2

Therefore

6.2moles of C4H10 will gives 4 moles of 24.8 moles of CO2

Using the ideal gas equation

PV=nRT

P= 1.0atm

V=?

n= 24.8mol.

R=0.08206atmL/molK

T=20+273=293

V= 24.8 × 0.08206 × 293

V= 596L

Therefore the volume of CO2 produced is 596L