Question

Assuming an efficiency of 49.50%, calculate the actual yield of magnesium nitrate formed from 130.7 g of magnesium and excess copper(II) nitrate. Mg+Cu(NO3)2⟶Mg(NO3)2+Cu actual yield:

Answer 1

Answer:

73.41 g

Explanation:

$Mg + Cu(NO_{3})_{3}   ---> Mg(NO_{3})_{3} + Cu$

First find number of moles of Mg

n(Mg)$= \frac{mass}{molar mass} = \frac{130.7 g}{24.3050 g/mol}= 5.377 mol$

Second, find the number of moles of magnesium nitrate produced by this number of moles of magnesium.

The molar ratio is one is to one (using stoichiometric coefficients from the balanced equation). So 5.377 mol of magnesium produces 5.377 mol of magnesium nitrate

Third, find what mass of magnesium nitrate that is

mass(Mg(NO3)2) = mol * molar mass

                            = 5.377 mol * 148.3148 g/mol

                            = 148.3148 g

But the system is inefficient. It only has an efficiency of 49.50 so what will be the actual, not theoretical yield.

actual yield = efficiency * theoretical yield

                   =  49.5% * 148.3148 g

                    = 73.41  g