Question

What is the percent yield if 50.0 g of Cr is formed from the reaction of 150.0 g of `Cr_2O_3` (MW = 152.0 amu) with excess Al?

Answer 1

Approx `50%` yield

Explanation:

`Cr_2O_3 + 2Al +Delta rarr 2Cr + Al_2O_3`

`"Moles of "Cr_2O_3` `=` `(150*g)/(151.99*g*mol^-1)` `~= 1*mol`

`"Moles of "Cr` `=` `(50*g)/(52.00*g*mol^-1)` `~= 1*mol`

This is a nasty question as it is trying to get you to ignore the given stoichiometry of the reaction, and report 100% yield. However, 100% yield would give `104*g` chromium metal, because the decomposition yields 2 mol of chromium per mol of `Cr_2O_3`.