For the reaction 2S(s) + 3O2(g) →2 SO3(s) 2.0 moles of sulfur are placed in a container with an excess of oxygen and 1.7 moles of SO3 are collected. What is the percentage yield?

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For the reaction 2S(s) + 3O2(g) →2 SO3(s) 2.0 moles of sulfur are placed in a container with an excess of oxygen and 1.7 moles of SO3 are collected. What is the percentage yield?

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Answer 1 · 2017-12-14T23:35:13

$85%$

Explanation:

From the reaction equation coefficients, we can understand that for every $2$ moles of sulfur, $2$ moles of $"SO"_3$ are created. So if $2.0$ moles of sulfur are placed in a container, where oxygen does not limit the reaction, $2.0$ moles of $"SO"_3$ are expected to be formed.

However, $1.7$ moles were collected experimentally. To show this disparity, we calculate the percentage yield, which is

$"% yield" = ("experimental value"/"actual value") * 100%$

So

$"% yield" = ("1.7 moles"/"2.0 moles")*100% = 85%$

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For the reaction 2S(s) + 3O2(g) →2 SO3(s) 2.0 moles of sulfur are placed in a container with an excess of oxygen and 1.7 moles of SO3 are collected. What is the percentage yield?

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Explanation:

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