If one pound (454g) of hexane combusts with 185 grams of oxygen which reactant is the limiting reactant and how many grams of carbon dioxide are produced?
My work: 454 g C6H14 * 1 mol C6H14/86.18 g C6H14 * 6 mol CO2/ 1 mol C6H14 * 44 g C02/ 1 mol CO2= 1390.7 g/ CO2
185 g O2 * 1 mol O2/ 32 g O2 * 6 mol CO2/(19/2) mol O2 * 44 .0g CO2/ 1 mol CO2= 160.65 g/CO2
Limiting factor is O2
The process looks ok to me but you have too many significant figures in the answer. I would round that to 161 grams.
To determine the limiting reactant and the amount of carbon dioxide produced, you need to calculate the amount of CO2 produced from each reactant and compare them.
First, let's convert the mass of hexane (C6H14) and oxygen (O2) into moles.
For hexane (C6H14):
454 g C6H14 * (1 mol C6H14 / 86.18 g C6H14) = 5.27 mol C6H14
For oxygen (O2):
185 g O2 * (1 mol O2 / 32 g O2) = 5.78 mol O2
Next, we calculate the amount of CO2 produced from each reactant.
For hexane (C6H14):
5.27 mol C6H14 * (6 mol CO2 / 1 mol C6H14) * (44 g CO2 / 1 mol CO2) = 1384.44 g CO2
For oxygen (O2):
5.78 mol O2 * (6 mol CO2 / 19/2 mol O2) * (44 g CO2 / 1 mol CO2) = 786.68 g CO2
From these calculations, we can see that the amount of CO2 produced from hexane is 1384.44 g, while the amount produced from oxygen is 786.68 g.
Since the amount produced from oxygen is lower, it indicates that oxygen is the limiting reactant. It is fully consumed, and the hexane is in excess.
Therefore, the limiting reactant is oxygen (O2), and the amount of carbon dioxide produced is 786.68 grams.