Question

Gaseous ethane CH3CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 3.86g of water is produced from the reaction of 6.92g of ethane and 39.7g of oxygen gas, calculate the percent yield of water. Round your answer to 3 significant figures.

Answer 1

CH3CH3 (g) + 7/2 O2 (g) -------> 2CO2 (g) + 3H2O (g)

No of moles of ethane = Mass/Molar mass of ethane => 6.92 g/ 30.07 gmol^-1 = 0.23 moles

No of moles of oxygen = Mass/molar mass of oxygen => 39.7 g/ 32 gmol^-1 => 1.24 moles

According to the reaction, 1 mole of ethane reacts with 3/2 mole of oxygen to form 3 moles of H2O.

Let x be the dissociated no of moles

If O2 is the limiting reagent then, no of moles of O2 = 0

=> 1.24 - (7/2)x = 0

=> x = 0.354

And no of moles of CH3CH3 = 0.23 - 0.354 = negative

Since, no of moles can not be negative, hence ethane is the limiting reagent.

Reaction	CH3CH3	7/2 O2	2 CO2	3 H2O
Initial moles	0.23	1.24	-	-
Moles after Reaction	0.23 - 0.23 => 0.23	1.24 - 7/2(0.23) => 0.435	2 * 0.23 => 0.46	3 * 0.23 => 0.69

Hence, no of moles of H2O formed = 0.69 moles

Now, Mass of H2O formed = No of moles * Molar Mass of H2O

=> Mass of H2O formed = 0.69 moles * 18 gmol^-1 => 12.42g

Thus, %yield = (actual yield/theoritical yield) * 100

=> %yield = ( 3.86 g / 12.42 g ) * 100

=> %yield = 31.08% (approx)

Hence the percentage yield of water is 31.08% (approx)