Question

In the reaction of 4.3 g of H₂ and 6.11 g of O₂, 2.4 g of water was isolated. what is the percent yield of the water?

*** PLEASE SHOW HOW YOU GET THE ANSWER***

Answer 1

The reaction is as follows

2H₂ + O₂ 2H₂O

So, 2 mole hydrogen will react with1 mole oxygen to produce 2 mole water

Then, 2 x 2 g hydrogen will react with 32 g oxygen to produce 2 x 18 g water (MW of O2 is 32)

So, 4 g hydrogen will react with 32 g oxygen to produce 36 g water

When , 4.3 g hydrogen is mixed with 6.11 g oxygen to produce water. then oxygen will be the limiting reagent which will consumed fully.

Now, 32 g oxygen react with 4 g hydrogen to produce  36 g water

So, 6.11 g oxygen will produce (36 x 6.11 / 32 ) = 6.8737 g water

Then theoritical yield of the reaction = 6.87 g of water

But, actual yield is 2.4 g water

So, percent yiled of the reaction = actual yield / theoritical yield x 100

= ( 2.4 / 6.87 ) x 100

= 34.93 % (ANSWER)