Question

When 1.89g Al is placed in 50.0 mL of an iron(II) solution, a reaction occurs producing solid metal iron and aluminum ions in solution. After the reaction, the aluminum had a mass of only 1.29 g. How much iron should have been produced? If all the iron reacted, what was the original concentration of the iron(II) solution? If only 1.55 g Fe were produced what was the percent yield?

Answer 1

2Al(S) + 3Fe^2+(aq)   ---> 2Al^3+(aq) + 3Fe(s)

no of mol of Al reacted = (1.89-1.29)/27 = 0.022 mol

from equation , 2 mol Al = 3 mol Fe

no of mol of Fe formed = 0.022*3/2 = 0.033 mol

mass of Fe formed = 0.033*56 = 1.848 g

theoretical yield of iron = 1.848 g

practical yield of iron = 1.55 g

percent yield = (practical yield )/(theoretical yield)*100

              = 1.55/1.848*100

       = 83.9%