Question

Assuming an efficiency of 25.10%, calculate the actual yield of magnesium nitrate formed from 146.2 g of magnesium and excess copper(II) nitrate. Mg + Cu(NO3)2 ----> Mg(NO3)2 +Cu Please explain your work

Answer 1

molar mass of Magnesium = 24.31 gm/mol

Number of moles of Magnesium = 146.2/24.31 = 6.0139 moles

Since excess Copper(II) nitrate is present, hence magnesium will be the limiting reagent

Now one mole of Mg gives 1 mole of Mg(NO3)2

Theoritical Yield of Mg(NO3)2 = 6.0139 moles

Molar mass of Mg(NO3)2 = 24.31 + (14 + 48)2 = 24.31 + 124 = 148.31 gm/mol

Theoritical yield = number of moles * molar mass = 6.0139 mol * 148.31 gm/mol = 891.93 gms

Actual Yield = 25.1% of Theoritical Yield = 25.10/100 * 891.93 = 223.875 grams