Question

Liquid octane CH3CH26CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . If 7.77g of water is produced from the reaction of 11.4g of octane and 68.8g of oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

PLEASE have the right number of significant digits

Answer 1

Sol:-

2C₈H₁₈ (l).......... + ........... 25O₂ (g) -------------> 16CO₂ (g)............... +............. 18H₂O (l)

2mol....................................25mol.........................16mol.....................................18mol......

228g...................................800g............................704g.......................................324g.........

Now from the equation it is cleared that

228g of octane reacts with = 800g of oxygen

so

11.4 g of octane reacts with = 800 g x 11.4g / 228 g = 40 g of oxygen .

because the given amount of oxygen is more i.e 68.8 g therefore oxygen is excessive reagent and octane is limiting reagent.

Now 228 g of octane gives = 324g of water

so

11.4 g of octane produces = 324 g x 11.4 g / 228 g = 16.2 g of water

therefore Theoretical yield = 16.2g of water

given Experimental yield = 7.77 g of water

so

Percent yield = Experimental yield x 100 / Theoretical yield

Percent yield = 7.77 g x 100 / 16.2g

Percent yield = 47.96 %

Percent yield = 48.0 %