Using the balanced chemical equation for the combustion of liquid dodecane into gaseous carbon dioxide and gaseous water, answer the following question. Suppose 0.380 kg of Dodecanese are burned in air at a pressure of exactly one atmosphere and a temperature of 11°C. Calculate the volume of carbon dioxide gas that is produced round your answer to three significant digits.
To solve this problem, we need to use the balanced chemical equation for the combustion of dodecane (C12H26):
2 C12H26 + 37 O2 -> 24 CO2 + 26 H2O
From the balanced equation, we can see that 2 moles of dodecane produce 24 moles of carbon dioxide gas. Thus, we need to find the moles of dodecane in 0.380 kg:
moles of dodecane = mass of dodecane / molar mass of dodecane
The molar mass of dodecane (C12H26) can be calculated by adding up the molar masses of carbon (C) and hydrogen (H):
Molar mass of C12H26 = (12*12.01) + (26*1.01) = 170.34 g/mol
Converting the mass of dodecane to grams:
mass of dodecane = 0.380 kg * 1000 g/kg = 380 g
moles of dodecane = 380 g / 170.34 g/mol = 2.232 mol (rounded to three significant digits)
Now, we can use the stoichiometric ratio from the balanced equation to find the moles of carbon dioxide produced:
moles of CO2 = (24 mol CO2 / 2 mol dodecane) * 2.232 mol dodecane = 26.91 mol (rounded to three significant digits)
Finally, we can use the ideal gas law to calculate the volume of carbon dioxide gas produced. The ideal gas law is given by:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Rearranging the equation to solve for V:
V = (nRT) / P
Given:
Pressure (P) = 1 atm
Temperature (T) = 11°C (or 11 + 273.15 = 284.15 K)
Number of moles (n) = 26.91 mol
Ideal gas constant (R) = 0.0821 L·atm/mol·K (rounded to three significant digits)
V = (26.91 mol * 0.0821 L·atm/mol·K * 284.15 K) / 1 atm = 604.25 L
The volume of carbon dioxide gas produced is 604.25 L (rounded to three significant digits).