Question

Assuming an efficiency of 34.50%, calculate the actual yield of magnesium nitrate formed from 133.3 g of magnesium and excess copper(II) nitrate.

Answer 1

The reaction we have is

Mg (s) + Cu(NO3)2 ----> Mg(NO3)2 + Cu (s)

since Copper nitrate is in excess Magnesium is limiting reagent

Mg moles = ( mass of Mg / molar mass of Mg) = ( 133.3g / 24.3 g/mol) = 5.486 mol

5.486 mol Mg reacts to give 5.486 mol Mg(NO3)2 if effeciency is 100 % ( as per reaction coeffcients)

considering 34.5 % efficiency Mg(NO3)2 moles we get = ( 34.5 /100) x 5.486 = 1.89 mol

Mg(NO3)2 mass = moles x molar mass of Mg(NO3)2 = 1.89 mol x 148.3 g/mol

= 281 g

Thus actuall yield of Magnesium nitrate is 281 grams