A common laboratory preparation for small for quantities of $O_2$ is to decompose $KClO_3$ by heating: $2KClO_3 (s) -> 2KCl (s) + 3O_2 (g)$. If the decomposition of 2.00 g of $KCIO_3$ gives 0.720 g of $O_2$, what is the percent yield for the reaction?

Question

A common laboratory preparation for small for quantities of $O_2$ is to decompose $KClO_3$ by heating: $2KClO_3 (s) -> 2KCl (s) + 3O_2 (g)$. If the decomposition of 2.00 g of $KCIO_3$ gives 0.720 g of $O_2$, what is the percent yield for the reaction?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

Start by examining the balanced chemical equation that describes this

$color(darkgreen)(2)"KClO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"KCl"_ ((s)) + color(blue)(3)"O"_ (2(g))$

Notice that it takes $color(darkgreen)(2)$ moles of potassium chlorate to produce $color(blue)(3)$ moles of oxygen gas. This represents the reaction's theoretical yield, i.e. the amount of oxygen gas that you can expect to see for a reaction that has a $100%$ .

The reaction provides you with grams of potassium chlorate, so use the molar masses of potassium chlorate and of oxygen gas to convert the mole ratio to a gram ratio.

$(color(darkgreen)(2)color(red)(cancel(color(black)("moles KClO"_3))) * "122.55 g"/(1color(red)(cancel(color(black)("mole KClO"_3)))))/(color(blue)(3)color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2))))) = "2.553 g KClO"_3/"1 g O"_2$

This means that for every $"2.553 g"$ of potassium chlorate that undergo combustion, the reaction can theoretically produce $"1 g"$ of oxygen gas.

Now, the problem tells you that you decompose $"2.00 g"$ of potassium chlorate. The reaction should theoretically produce

$2.00 color(red)(cancel(color(black)("g KClO"_3))) * "1 g O"_2/(2.553color(red)(cancel(color(black)("g KClO"_3)))) = "0.7834 g O"_2$

This represents the reaction's theoretical yield. You know that the reaction actually produced $"0.720 g"$ of oxygen gas, which means that its was

$(0.720 color(red)(cancel(color(black)("g O"_2))))/(0.7834color(red)(cancel(color(black)("g O"_2))))xx 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(91.9%)color(white)(a/a)|)))$

The answer is rounded to three .

A common laboratory preparation for small for quantities of $O_2$ is to decompose $KClO_3$ by heating: $2KClO_3 (s) -&gt; 2KCl (s) + 3O_2 (g)$. If the decomposition of 2.00 g of $KCIO_3$ gives 0.720 g of $O_2$, what is the percent yield for the reaction?

1 Answers

A common laboratory preparation for small for quantities of $O_2$ is to decompose $KClO_3$ by heating: $2KClO_3 (s) -> 2KCl (s) + 3O_2 (g)$. If the decomposition of 2.00 g of $KCIO_3$ gives 0.720 g of $O_2$, what is the percent yield for the reaction?