Question

Combustion of hydrocarbons such as decane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide. 1. Write a balanced chemical equation, including physical state symbols, for the combustion of liquid decane into gaseous carbon dioxide and gaseous water. 2. Suppose of decane are burned in air at a pressure of exactly and a temperature of . Calculate the volume of carbon dioxide gas that is produced. Round your answer to significant digits.

Answer 1

The missing part of the question is shown in the image attached

Answer:

C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)

V= 70.4L of CO2

Step-by-step explanation:

Equation of the reaction is:

C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)

Number of moles of decane = mass/ molar mass

Molar mass of decane= 122gmol-1

n= 0.370×10^3g/122gmol-1= 3.0 moles

T= 13°C +273=286K

P= 1atm

R= 0.082 atmLK-1mol-1

From :

PV= nRT

V= nRT/P

V= 3.0×0.082×286/1

V= 70.4L of CO2