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 moles of propane are formed? Be sure your answer has the correct number of significar
mol C3H8
First, we need to calculate the moles of propene (C3H6) and hydrogen (H2) used in the reaction:
Moles of C3H6 = 6.00 g / 42.08 g/mol = 0.143 mol
Moles of H2 = 16.00 g / 2.016 g/mol = 7.94 mol
Next, we determine the limiting reactant by comparing the moles of C3H6 and H2:
From the balanced equation, we see that 1 mole of C3H6 reacts with 1 mole of H2 to produce 1 mole of propane (C3H8). Therefore, the limiting reactant is the one that produces the least amount of propane.
Since 0.143 mol of C3H6 produces 0.143 mol of C3H8 and 7.94 mol of H2 produces 7.94 mol of C3H8, the limiting reactant is C3H6.
So, the moles of propane (C3H8) formed will be equal to the moles of C3H6 used, which is 0.143 mol of C3H8.