Question

Assuming an efficiency of 27.80%, calculate the actual yield of magnesium nitrate formed from 122.4 g of magnesium and excess copper(II) nitrate.

Mg+ Cu(NO3)2 = Mg(NO3)2 + Cu

Answer 1

we know that

moles = mass / molar mass

so

moles of Mg = 122.4 / 24 = 5.1

now

consider the given reaction

Mg + Cu(N03)2 ---> Mg(N03)2 + Cu

we can see that

theoretical moles of Mg(N03)2 = moles of Mg reacted

so

theoretical moles of Mg(N03)2 = 5.1

now

we know that

mass = moles x molar mass

so

theoretical mass of Mg(N03)2 = 5.1 x 148

theoretical mass of Mg(N03)2 = 754.8

now

we know that

% yield = actual x 100 / theoretical

so

27.8 = actual x 100 / 754.8

actual = 210

so

210 grams of Mg(N03)2 is the actual yield