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?

1 Answer
Dec 14, 2017

Answer:

#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%#