Final answer:
When heptane is combusted, it reacts with oxygen to produce carbon dioxide and water. The balanced chemical equation for this reaction is C7H16 (l) + 11O2 (g) --> 7CO2 (g) + 8H2O (g). To calculate the volume of carbon dioxide gas produced when 0.140 kg of heptane is burned, we convert the mass of heptane to moles and then use the ideal gas law to calculate the volume.
Step-by-step explanation:
When heptane (C7H16) is combusted, it reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced chemical equation for this reaction is:
C7H16 (l) + 11O2 (g) → 7CO2 (g) + 8H2O (g)
To calculate the volume of carbon dioxide gas produced when 0.140 kg of heptane is burned, we first need to convert the mass of heptane to moles. The molar mass of heptane is 100.20 g/mol. So, we have:
0.140 kg × (1000 g/1 kg) × (1 mol/100.20 g) = 1.396 mol of heptane
According to the balanced equation, 1 mol of heptane produces 7 mol of carbon dioxide gas. So, 1.396 mol of heptane will produce:
1.396 mol × 7 mol/mol = 9.772 mol of carbon dioxide gas
Finally, we can use the ideal gas law to calculate the volume of the carbon dioxide gas. The ideal gas law equation is:
V = nRT/P
Where V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), T is the temperature in Kelvin, and P is the pressure in atmospheres.
Since we are given the temperature in degrees Celsius, we need to convert it to Kelvin. The conversion formula is:
T(K) = T(°C) + 273.15
So, we have:
T(K) = 11.0 °C + 273.15 = 284.15 K
Plugging in the values, we get:
V = 9.772 mol × (0.0821 L·atm/(mol·K)) × 284.15 K / 1 atm = 226.221 L
Therefore, the volume of carbon dioxide gas produced is approximately 226.221 L.