If i react 2.45 mol H2 with 1.89 mol O2 and I obtain 1.56 mol of H2O2, what is t

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 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
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

Explanation / Answer

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 %

  

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