Question

Combustion of hydrocarbons such as undecane (C_11H_24) 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 undecane into gaseous carbon dioxide and gaseous water.
2. Suppose 0.260 kg of undecane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0°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:

Volume of carbon dioxide is 428.23 L.

Explanation:

Below is the chemical reaction or chemical equation for the combustion of hydrocarbons such as undecane into carbon dioxide.

$C_{11}H_{24}(l) + 17O_{2}(g) \to 11CO_{2} (g) + 12H_2O(g)$

Here, undecane is in liquid form that reacts with gaseous oxygen (combustion)  and produces carbon dioxide and water as a product in the gaseous form.

The molar mass of undecane = $156.31 g/mol.$

$\text{Number of moles} = \frac{260g}{156.31g/mol} = 1.66 \ mol.$

From the equation, it can be seen that 1 mole of undecane produces 11 moles of carbon dioxide. Therefore, 1.66 mol will produce 18.26 mol of carbon dioxide.

Now find the volume of 18.26 mol of carbon dioxide when the temperature is 13 degrees Celsius and pressure is 1 atm.

$V = \frac{nRT}{P}$

$V = \frac{18.26 \times 0.082 \times 286 \ K}{1atm}$

$V = 428.23 \ L$