Propane (C3H8, molar mass=44.09g/mol) can be prepared by the reaction of propene C3H6 molar mass=42.08 g/mol) with hydrogen (H2, molar mass=2.016 g/mol) according to the following balanced equation: C3H6+H2 C3H8
Assuming18.50g of propene and 28.50 of hydrogen are used, answer the following questions:
How many moles of propene are used? Be sure your answer has the correct number of significant figures.
How many grams of propane are formed? Be sure your answer has the correct number of significant figures.
To find the number of moles of propene used, we can use the molar mass of propene.
Molar mass of propene (C3H6) = 42.08 g/mol
Given mass of propene used = 18.50 g
Number of moles of propene used = mass of propene used / molar mass of propene = 18.50 g / 42.08 g/mol
Calculating the result gives:
Number of moles of propene used = 0.4399 mol
Therefore, 0.4399 moles of propene are used.
To find the mass of propane formed, we can use the molar mass of propane.
Molar mass of propane (C3H8) = 44.09 g/mol
Using the stoichiometry of the balanced equation, we know that 1 mole of propene reacts to form 1 mole of propane.
Number of moles of propane formed = number of moles of propene used = 0.4399 mol
Mass of propane formed = number of moles of propane formed * molar mass of propane = 0.4399 mol * 44.09 g/mol
Calculating the result gives:
Mass of propane formed = 19.38 g
Therefore, 19.38 grams of propane are formed.