For the following chemical equation, a student reacts 115 grams of pentane (C5H12)with 325 grams of oxygen. Calculate the mass of pentane and the mass of oxygen (in grams) that remained after the reaction was complete. Also,if the reaction produces a total of 2575 kJ of energy, calculate the percent yield.
C5H12(g) + 8 O2(g)5 CO2(g) + 6 H2O(g) + 3525kJ
I) To know the mass of oxygen and pentane that remained after the reaction was completed, let's find the limiting reagent that was used up first.
From the balanced equation,
72g of pentane reacted with 256g of oxygen
115g of pentane should react with 256/72 x 115=409g of oxygen
But only 325g of oxygen was available, oxygen was therefore the limiting reagent and it was completely used up before the reaction was completed.
Now, let's find the mass of pentane that reacted with 325g of O2
256g of O2 reacted with 72g of pentane
325g of oxygen would react with 72/256 x 325 = 91g
Mass of pentane remaining= 115-91= 24g of pentane
Mass of oxygen remaining= 0g of oxygen
II) for the percentage yield of energy
%yield= experimental yield/theoretical yield x 100%
= 2575KJ/3525KJ x 100%=73%