Calculation This is a limiting reagent problem based on the

Calculation: This is a limiting reagent problem, based on the following equation: CuSO4 . 5H20+ Fe Feso, Cu + 5H2O + Use the above described method of \"Limiting Reagent.\" Calculate the yield of Cu, and compare it with the experimental value from step 11.

Solution

i think you have not given full information . so i am assuming values by myself.

Let 4 g of copper sulfate pentahydrate and 0.5 g of iron

Then no of moles opper sulfate pentahydrate = 4g CuSO4·5H2O / (249.72 g/mol CuSO4·H2O) = 0.016 mol

No of moles of iron = 0.5 g Fe / (55.85 g/mol Fe) = 0.009 mol

Using molar ratio :Cu / CuSO4.5H2O
from the equation: 1mol Cu / 1 mol CuSO4.5H20
using your result : 1mol Cu/ 1mol CuSO4.5H2O * 0.016mol CuSO4.5H2O =0.016 mol CU

Fe :
Using molar ratio : Cu / Fe
from the equation : 1mol Cu / 1mol Fe
using your result : 1mol Cu / 1mol Fe * 0.009 mol Fe =0.009 mol Cu

Fe is the limiting reagent.