Question

Suppose 0.210kg of octane 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. Round your answer to 3 significant digits.

Answer 1

Answer: The volume of carbon dioxide gas that is produced from the given amount of octane is $3.29\times 10^2L$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of octane = 0.210 kg = 210 g     (Conversion factor: 1 kg = 1000 g)

Molar mass of octane = 114.23 g/mol

Putting values in above equation, we get:

$\text{Moles of octane}=\frac{210g}{114.23g/mol}=1.84mol$

The temperature and pressure conditions given to us are NTP conditions:

At NTP:

1 mole of a gas occupies 22.4 L of volume.

So, 1.84 moles of octane gas will occupy $22.4\times 1.84=41.216L$ of volume.

The chemical reaction for the combustion of octane follows the equation:

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

By Stoichiometry of the reaction:

$(2\times 22.4L)$ of octane gas produces $(16\times 22.4L)$ of carbon dioxide gas.

So, 41.216 L of octane gas will produce = $\frac{(16\times 22.4)L}{(2\times 22.4)L}\times 41.216=329.728L=3.29\times 10^2L$ of carbon dioxide gas.

Hence, the volume of carbon dioxide gas that is produced from the given amount of octane is $3.29\times 10^2L$

Answer 2

Answer:

0.708 L

Explanation:

Start with the balanced chemical equation:

$2C_{8}H_{18} +25O_{2} --->16CO_{2} +18H_{2}O$

calculate the number of moles of octane:

n(octane)=$\frac{mass}{molar mass}=\frac{0.210g}{114.23g/mol}=1.8384*10^{-3}mol$

From the balanced equation we see that the number of carbon dioxide produced is always 16 times the number of moles of octane available.

Therefore n(carbon dioxide)=$1.8384*10^{-3} mol *16=0.0294144mol$

We now treat CO2 as an idal gas, which in reality it behaves like. That allows us to use the following ideal gas equation.

$Volume=\frac{nRT}{P}=\frac{0.0294144mol*0.08205L.atm/(mol.K)*293.15K}{1 atm} =0.7075L$