$2H_2O -> 2H_2 + O_2$ 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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You know by looking at the balanced chemical equation

$2"H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O"_ (2(g))$

that it takes $2$ moles of water to produce $1$ mole of oxygen gas. Use the molar masses of the two chemical species to convert this mole ratio to a gram ratio

$("2 moles H"_2"O")/("1 mole O"_2) = (2 color(red)(cancel(color(black)("moles H"_2"O"))) * ("18.015 g H"_2"O")/(1color(red)(cancel(color(black)("mole H"_2"O")))))/(1color(red)(cancel(color(black)("mole O"_2))) * "32.0 g O"_ 2/(1color(red)(cancel(color(black)("mole O"_2))))) = ("36.03 g H"_2"O")/("32.0 g O"_2)$

So, you know that every $"36.03 g"$ of water that take part in the reaction produce $"32.0 g"$ of oxygen gas.

This means that when $"17.0 g"$ of water undergo decomposition, you should expect to get

$17.0 color(red)(cancel(color(black)("g H"_2"O"))) * "32.0 g O"_2/(36.03color(red)(cancel(color(black)("g H"_2"O")))) = "15.1 g O"_2$

This represents the reaction's theoretical yield, i.e. what is produced by a reaction that has a $100%$ yield.

Now, you know that the actual yield of the reaction is $"10.2 g"$ of oxygen gas. To find the , divide the actual yield by the theoretical yield and multiply the result by $100%$.

You should get

$"% yield" = (10.2 color(red)(cancel(color(black)("g O"_2))))/(15.1color(red)(cancel(color(black)("g O"_2)))) * 100% = color(darkgreen)(ul(color(black)(67.5%)))$

The answer is rounded to three .

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