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?
how do I do this? How do I set it up?
Sarah, a stoichiometry problem is a stoichiometry problem. See your last post with hexane. Post what you know to do and explain in detail what you don't understand next.
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
To determine the limiting reactant and the amount of carbon dioxide produced in this reaction, you need to follow these steps:
1. Write a balanced chemical equation for the combustion of hexane (C6H14) with oxygen (O2):
C6H14 + 19O2 -> 6CO2 + 7H2O
2. Convert the given masses of hexane (454 grams) and oxygen (185 grams) into moles. To do this, divide the mass by the molar mass of each substance.
Molar mass of hexane (C6H14) = 86.18 g/mol
Molar mass of oxygen (O2) = 32.00 g/mol
Moles of hexane = 454 g / 86.18 g/mol = 5.27 mol
Moles of oxygen = 185 g / 32.00 g/mol = 5.78 mol
3. Calculate the stoichiometric ratio between hexane and oxygen. Since the balanced equation shows that 1 mole of hexane reacts with 19 moles of oxygen, divide the moles of oxygen by 19 to find the ratio.
Ratio of moles of oxygen to hexane = 5.78 mol / 19 = 0.304 moles of oxygen per mole of hexane
4. Compare the stoichiometric ratio with the actual ratio. The actual ratio is 5.27 moles of hexane to 5.78 moles of oxygen, which is approximately 0.913 moles of oxygen per mole of hexane.
5. The reactant with the smaller ratio (0.304 moles) is the limiting reactant. In this case, hexane is the limiting reactant.
6. Determine the amount of carbon dioxide produced. Since 1 mole of hexane produces 6 moles of carbon dioxide, you can use the mole ratio to calculate the moles of carbon dioxide produced:
Moles of carbon dioxide = moles of hexane x (6 moles of CO2/1 mole of hexane)
Moles of carbon dioxide = 5.27 mol x 6 = 31.62 mol
7. Finally, convert moles of carbon dioxide into grams by multiplying by the molar mass of carbon dioxide (44.01 g/mol):
Grams of carbon dioxide = moles of carbon dioxide x molar mass of carbon dioxide
Grams of carbon dioxide = 31.62 mol x 44.01 g/mol = 1390.84 g
Therefore, in the combustion of one pound (454g) of hexane with 185 grams of oxygen, the limiting reactant is hexane and 1390.84 grams of carbon dioxide are produced.