Question

You carry out the reaction as described below, using 16.2 g of copper, 81.3 mL of 3.0 M H2SO4, and 20.8 mL of 16 M HNO3. What is the limiting reagent in this reaction?

Cu(s) + 2HNO3(aq) + H2SO4(aq) + 3H2O ---> CuSO4.5H2O(s) + 2NO2(g)

What is the theoretical yield of CuSO4·5H2O in grams?

Answer 1

No. of moles of Copper = 16.2 / 63.546 = 0.2549 moles

No. of moles of H2SO4 = 3.0 x 0.0813 = 0.2439 mole

no. of moles of HNO3 = 16 x 0.0208 = 0.3328 moles

from the reaction; 1 mole of Cu and 2 moles of HNO3 and 1 mole of H2SO4 and excess water will react to give product.

Actual HNO3 reacted = (1/2) x 0.3328 = 0.1664 mole

For limiting reagent one which is present in lower no. of moles will be the limiting reagent.

So, limiting reagent is HNO3

limiting no. of moles = 0.1664

Actual reacted reactants are Cu = 0.1664 moles, H2SO4 = 0.1664 moles and HNO3 = 2 x 0.1664 will give product is = 0.1664 moles

Molecular mass of product = 249.677

Weight of product formed (theoretical yield) = 0.1664 x 249.677 = 41.55 grams