C3H8, molar mass = 44.09
g
mol
C3H6 + H2 + C3H8
Assuming 6.00 g of propene and 16.00 g of hydrogen are used, answer the following questions:
Propane
H2, molar mass = 2.016
g
mol can be prepared by the reaction of propene
C3H6, molar mass = 42.08
g
mol
with hydrogen
according to the following balanced equation: moles of propane =
How many grams of propane are formed? Be sure your answer has the correct number of significant figures.
mass of propane =
g C3H8
To determine how many grams of propane (C3H8) are formed, we first need to find the limiting reactant between propene (C3H6) and hydrogen (H2).
1. Calculate the number of moles of propene (C3H6):
moles of C3H6 = 6.00 g / 42.08 g/mol = 0.1426 mol
2. Calculate the number of moles of hydrogen (H2):
moles of H2 = 16.00 g / 2.016 g/mol = 7.9375 mol
3. Determine the limiting reactant by comparing the stoichiometry of the balanced equation:
C3H6 + H2 -> C3H8
From the balanced equation, for every 1 mol of C3H6, 3 moles of H2 are required. Therefore, the actual moles of H2 required should be 0.1426 mol x 3 = 0.4278 mol.
Since the actual moles of H2 (0.4278 mol) are less than the moles of H2 provided (7.9375 mol), hydrogen is in excess and propene is the limiting reactant.
4. Calculate the maximum number of moles of propane (C3H8) that can be formed based on the limiting reactant (propene):
moles of C3H8 = 0.1426 mol
5. Calculate the mass of propane in grams:
mass of C3H8 = moles of C3H8 x molar mass of C3H8
mass of C3H8 = 0.1426 mol x 44.09 g/mol = 6.28 g
Therefore, 6.28 grams of propane (C3H8) can be formed. This answer has the correct number of significant figures.