Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 43.9 g of hexane is mixed with 57. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Round your answer to significant digits.
To determine the minimum mass of hexane that could be left over by the chemical reaction, we need to calculate the limiting reactant. The reactant that is completely consumed in the reaction will determine the amount of product formed.
First, let's convert the given masses of hexane and oxygen into moles by using their respective molar masses.
The molar mass of hexane (C6H14) is:
(6 x atomic mass of carbon) + (14 x atomic mass of hydrogen)
= (6 x 12.01 g/mol) + (14 x 1.01 g/mol)
= 72.06 g/mol + 14.14 g/mol
= 86.20 g/mol
Now, we can calculate the moles of hexane:
moles of hexane = mass of hexane / molar mass of hexane
moles of hexane = 43.9 g / 86.20 g/mol
moles of hexane = 0.509 mol
Similarly, let's calculate the moles of oxygen. The molar mass of oxygen is approximately:
(2 x atomic mass of oxygen)
= (2 x 16.00 g/mol)
= 32.00 g/mol
moles of oxygen = mass of oxygen / molar mass of oxygen
moles of oxygen = 57.0 g / 32.00 g/mol
moles of oxygen = 1.78 mol
Now, let's determine the limiting reactant. The balanced chemical equation for the reaction is:
2 C6H14 + 19 O2 -> 12 CO2 + 14 H2O
The stoichiometric ratio of hexane to oxygen is 2:19. This means that for every 2 moles of hexane, we need 19 moles of oxygen.
Comparing the moles of hexane (0.509 mol) and oxygen (1.78 mol), we see that the hexane is in excess. Therefore, the oxygen is the limiting reactant.
Now, let's calculate the mass of hexane that could be left over by the chemical reaction.
moles of hexane left over = moles of hexane - (moles of oxygen x stoichiometric ratio of hexane to oxygen)
moles of hexane left over = 0.509 mol - (1.78 mol x (2/19))
moles of hexane left over = 0.509 mol - (1.78 mol x 0.1053)
moles of hexane left over = 0.509 mol - 0.1874 mol
moles of hexane left over = 0.321 mol
Finally, let's convert the moles of hexane left over to mass:
mass of hexane left over = moles of hexane left over x molar mass of hexane
mass of hexane left over = 0.321 mol x 86.20 g/mol
mass of hexane left over = 27.69 g
Rounding to the correct number of significant digits, the minimum mass of hexane that could be left over by the chemical reaction is 27.7 g.
To calculate the minimum mass of hexane that could be left over, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, thus determining the maximum amount of product formed.
First, we need to convert the given masses of hexane and oxygen into moles.
The molar mass of hexane (C6H14) can be calculated as follows:
C: 6 x 12.01 g/mol = 72.06 g/mol
H: 14 x 1.01 g/mol = 14.14 g/mol
Adding the mass of carbon and hydrogen together:
72.06 g/mol + 14.14 g/mol = 86.20 g/mol
Now we can calculate the number of moles of hexane:
43.9 g / 86.20 g/mol = 0.509 mol
Next, we calculate the number of moles of oxygen:
57.0 g / 32.00 g/mol = 1.78 mol
Using the balanced chemical equation, we can determine the mole ratio between hexane and oxygen from the coefficients:
2 C6H14 + 19 O2 → 12 CO2 + 14 H2O
From the balanced equation, we see that 2 moles of hexane react with 19 moles of oxygen.
Now, let's calculate the mole ratio of hexane to oxygen:
0.509 mol hexane / 2 = 0.2545 mol hexane
1.78 mol oxygen / 19 = 0.0937 mol oxygen
The mole ratio tells us that for every 2 moles of hexane, we need 19 moles of oxygen. However, we have a higher amount of oxygen (1.78 mol) compared to the hexane (0.509 mol), which means that oxygen is in excess.
To find the amount of hexane that reacts completely, we need to use the mole ratio between hexane and oxygen (2:19) to determine the amount of hexane that reacts with the available oxygen.
Using the mole ratio, we find that the amount of hexane reacting with 1.78 mol of oxygen is:
0.0937 mol oxygen × (2 mol hexane / 19 mol oxygen) = 0.0098 mol hexane
Finally, we convert the moles of hexane to mass using the molar mass:
0.0098 mol × 86.20 g/mol = 0.845 g hexane
Therefore, the minimum mass of hexane that could be left over is 0.845 grams.
This is a limiting reagent (LR) problem and I do these the long way.
Write and balance the equation.
2C6H14 + 19O2 ==>12CO2 + 14H2O
Convert 57 g O2 to mols. mol = grams/molar mass = ?
Do the same to convert 43.9 g hexane to mols.
Using the coefficients in the balanced equation, convert mols hexane to mols CO2
Do the same to convert mols O2 to mols CO2.
It is likely that these two values will not agree which means one is wrong. The correct value in LR problem is ALWAYS the smaller value and the reagent responsible for producing that value is the LR.
After you know the LR, you can determine how much of the excess reagent (the one that isn't the LR). Again, use the coefficients in the balanced equation to convert mols of the LR to mols of the excess reagent. Then to find how much is left, it is
initial amount - amount needed to react with the LR = amount in excess.
Post your work if you get stuck.