The combustion of 75.5 liters of 2- propane (c3h6)with 95.5 liters of oxygen gas is extremely exothermic. What is the maximum amount of water (in grams) to be made from this and how much of the non-limiting reactant is unused at the end (in liters)?
The equation is (note no lower case letters used)
2C3H6 + 9O2 ==> 6CO2 + 6H2O
Convert L C3H6 to L CO2.
75.5L x (6 mols CO2/2 moles C3H6) = 226.5 L CO2.
Convert 95.5 L O2 to L CO2.
95.5 x (6 moles CO2/9 moles O2) = 63.7 L CO2.
In limiting reagent problems, the smaller value of the product formed is ALWAYS the correct value; therefore, O2 is the limiting reagent.
This should serve as an example; you can do it for H2O. Use the same kind of process to determine how much of the C3H6 was used, then subtract from the initial amount to determine the amount not reacted.
To solve this problem, we need to determine the limiting reactant, which is the reactant that is completely consumed and limits the amount of product formed.
To find the limiting reactant, we can compare the amount of each reactant to the stoichiometric ratio of the balanced chemical equation. The balanced equation for the combustion of propane (C3H8) is:
C3H8 + 5O2 -> 3CO2 + 4H2O
According to the equation, we need 5 moles of oxygen gas (O2) for every 1 mole of propane (C3H8) to react completely.
Converting the given volumes of reactants (75.5 liters of propane and 95.5 liters of oxygen) to moles using the ideal gas law:
n = PV / RT
where:
n = number of moles
P = pressure
V = volume
R = gas constant
T = temperature in Kelvin
Let's assume the temperature and pressure are constant for both gases. The gas constant R is approximately 0.0821 L·atm/(mol·K).
For propane:
n(propane) = (75.5 L) / (0.0821 L·atm/(mol·K) * T)
n(propane) = (75.5 L) / (0.0821 L·atm/(mol·K))
For oxygen:
n(oxygen) = (95.5 L) / (0.0821 L·atm/(mol·K) * T)
n(oxygen) = (95.5 L) / (0.0821 L·atm/(mol·K))
Now, we can determine the limiting reactant by comparing the moles of each reactant to the stoichiometric ratio of the balanced equation.
For propane:
moles(propane) = n(propane)
For oxygen:
moles(oxygen) = n(oxygen) / 5
Compare the moles of each reactant:
moles(propane) / 1 = moles(oxygen) / 5
If the moles of propane divided by 1 are less than the moles of oxygen divided by 5, then propane is the limiting reactant. Otherwise, oxygen is the limiting reactant.
After finding the limiting reactant, we can determine the maximum amount of water produced by using stoichiometry.
If propane is the limiting reactant (moles(propane) / 1 < moles(oxygen) / 5):
moles(water) = 4 * moles(propane)
If oxygen is the limiting reactant (moles(propane) / 1 >= moles(oxygen) / 5):
moles(water) = 4 * (moles(oxygen) / 5)
Finally, we can convert the moles of water to grams using the molar mass of water (18.015 g/mol) and calculate the amount of unused reactant.
mass(water) = moles(water) * molar mass of water
To find the amount of the non-limiting reactant unused at the end, we need to subtract the amount of reactant used from the initial amount. We convert the moles of unused reactant to liters using the ideal gas law and the given temperature and pressure assumption.
Let's calculate step by step. But first, we need to know the temperature and pressure at which the reaction is taking place.