# 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?

Nov 21, 2015

%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 $10.2 g$.

We will need to find the theoretical yield.

The decomposition reaction of water can be written as:

$2 {H}_{2} O \left(l\right) \to 2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right)$

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 $15.1 g$.

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