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 $H_2O$?

Md Ariful Islam

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Md Ariful Islam

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The 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: $%Yield=(10.2)/(15.1)xx100%=67.5%$

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