Answer to Question #212786 in General Chemistry for jess

Question

50.0 g of C3H8 reacts with 50.0 g of O2 to produce CO2 and H2O. The limiting reactant is O2. The equation for the reaction is

C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)

a) What is the significance of the limiting reagent in predicting the amount of the product obtained in a reaction?

b) Calculate the theoretical yield (in grams) of water (H2O) that would form according to the above chemical reaction.

c) If the percentage yield of water (H2O) in the above reaction is 88.5 %, calculate the experimental yield of water (in grams).

d) Calculate the moles and mass of excess reactant remaining at the end of the reacti

Answer 1

The limiting reactant will be completely consumed in the reaction and limits the amount of product you can make. The limiting reactant also determines the amount of product you can make (the theoretical yield). The reactant that is left over after the reaction is complete is called the excess reactant.

Moles of C3H3$=\frac{50}{44.1}=1.134 moles$

Moles of O2​=1650​=3.125moles

Moles of $H2O = 1.134×4=4.536moles$

Or

H2O moles = $\frac{3.125}{5}×4=2.5moles$