If 7.89 g of water were recovered from the reaction, what is the % yield?

Question

20 g of octane are combusted with 2,400 ml of oxygen. If 7.89 g of water were recovered from the reaction, what is the % yield?

Any explanation or show of work so that I could understand would be much appreciated. Thanks!

Halchemist · Accepted Answer

C8H18  +  12.5 O2  ---------->  8 CO2(g)  +  9 H2O
 114 g........280 L.............................................162 g

? g H2O = 2.40 L O2 x 162 g H2O / 280 L O2 = 1.39 g H2O.
? g H2O = 20. g C8H18 x 162 g H2O / 114 g octane = 28.4 g H2O
 the above calculations it show oxygen to be the limiting reagent so a calculation for the % yield is impossible. Perhaps the yield should be 0.789 g: then the % yield is:
100% x 0.789 g/1.39 g = 56.7%

Trending News

If 7.89 g of water were recovered from the reaction, what is the % yield?

1 Answer