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?

1 Answer
Aug 25, 2016

Answer:

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#.