if the density of octane is 0.703 g/mol, how many gras of carbon dioxide are produced in the combustion of 40L of octane??
thats suppose to say grams!! not grass
are you from QUT? because that question is on the assignment
To determine the amount of carbon dioxide produced in the combustion of octane, we need to consider the balanced chemical equation for the combustion reaction. The balanced equation for the combustion of octane is as follows:
2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O
From the equation, we can see that for every 2 moles of octane, we produce 16 moles of carbon dioxide.
Given that the density of octane is 0.703 g/mol, we can calculate the number of moles of octane in 40 L of octane using the density and molar mass of octane. The molar mass of octane (C8H18) is calculated as follows:
12 g/mol (for carbon) × 8 + 1 g/mol (for hydrogen) × 18 = 114 g/mol
Using the density, we can calculate the number of moles of octane as follows:
40 L × (0.703 g/mol) / (114 g/mol) = 0.246 mol
Since the mole ratio between octane and carbon dioxide is 2:16, we can determine the number of moles of carbon dioxide produced:
0.246 mol (octane) × (16 mol CO2 / 2 mol octane) = 1.968 mol CO2
Finally, to find the mass of carbon dioxide produced, we multiply the number of moles by the molar mass of carbon dioxide (CO2), which is approximately 44 g/mol:
1.968 mol CO2 × 44 g/mol = 86.592 g CO2
Therefore, in the combustion of 40 L of octane, approximately 86.592 grams of carbon dioxide are produced.