Question

1. Write a balanced chemical equation, including physical state symbols, for the combustion of liquid dodecane into gaseous carbon dioxide and gaseous water.
2. Suppose 0.22 kg of dodecane are burned in air at a pressure of exactly and a temperature of 17.0°C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to significant digits.

Answer 1

Answer:  $2C_{12}H_{26}(l)+37O_2(g)\rightarrow 24CO_2(g)+26H_2O(g)$

2.  369 L

Explanation:

1. Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.

The balanced chemical reaction is:

$2C_{12}H_{26}(l)+37O_2(g)\rightarrow 24CO_2(g)+26H_2O(g)$

2. To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{220g}{170.3g/mol}=1.29moles$

According to stoichiometry :

2 mole of dodecane produces = 24 moles of carbon dioxide

Thus 1.29 moles of dodecane produces = $\frac{24}{2}\times 1.29=15.5 moles$ of carbon dioxide

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm

V = Volume of gas = ?

n = number of moles = 15.5

R = gas constant =$0.0821Latm/Kmol$

T =temperature =$17^0C=(17+273)K=290K$

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

$V=\frac{15.5\times 0.0821L atm/K mol\times 290K}{1atm}=369L$

Thus the volume of carbon dioxide gas that is produced is 369 L