Question

If i react 2.45 mol H2 with 1.89 mol O2 and I obtain 1.56 mol of H2O2, what is the percent yield of H2O?

The unbalanced equation is shown below:

H2 (g) + O2 (g) ---> H2O (g)

A). 82.5%

B). 77.1 %

C). 63.7%

D). 31.8%

E). 41.3%

please show me how you got this

Answer 1

H2 (g) + O2 (g) ---> H2O (g) Unbalanced

2H2 (g) + O2 (g) ---> 2H2O (g) balanced

GIven that 2.45 mol H2 reacted with with 1.89 mol O2.

O2 is the limiting reagent because no of moles of O2 is less.

Then,

2H2 (g) + O2 (g) ---> 2H2O (g)

1 mol 2 mol

  1.89 mol ? = 2 x 1.89 = 3.78 mol

This is theoretical yield of H2O.

But , you got 1.56 mol of H2O. This is actual yield of H2O

Hence,

percent yield of H2O =  actual yield of H2O / theoretical yield of H2O x 100

= 1.56 mol/ 3.78 mol x 100

= 41.3 %

Therefore,

percent yield of H2O = 41.3 %