Question

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?

Answer 1

`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%`