Question

What is the percent yield of `O_2` if 10.2 g of `O_2` is produced from the decomposition of 17.0 g of `H_2O`?

Answer 1

`%Yield=67.5%`

Explanation:

The percent yield can be calculated by:

`%Yield=("Actual Yield")/("Theoretical Yield")xx100%`

The actual yield of oxygen is given and it is `10.2g`.

We will need to find the theoretical yield.

The decomposition reaction of water can be written as:

`2H_2O(l)->2H_2(g)+O_2(g)`

To find the theoretical yield of oxygen we will use dimensional analysis:

`?g O_2=underbrace(17.0gH_2Oxx(1molH_2O)/(18.0gH_2O))_(color(blue)("g to mol"))xxunderbrace((1molO_2)/(2molH_2O))_color(green)("molar ratio")xxunderbrace((32.0gO_2)/(1molO_2))_(color(red)("mol to g"))=15.1gO_2`

Thus, the theoretical yield is equal to `15.1g`.

The percent yield is then: `%Yield=(10.2)/(15.1)xx100%=67.5%`