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#?

1 Answer
Nov 21, 2015

Answer:

#%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%#