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?

Dec 14, 2017

85%

Explanation:

From the reaction equation coefficients, we can understand that for every $2$ moles of sulfur, $2$ moles of ${\text{SO}}_{3}$ 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 ${\text{SO}}_{3}$ 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%