Question

Barium sulfate is made by the following reaction:

Ba(NO₃)₂ (aq) + Na₂SO₄ (aq) ? BaSO₄ (s) + 2NaNO₃ (aq)

An experiment was begun with 75.00 g of barium nitrate and 72.00g of sodium sulfate. After collecting and drying 64.45 g of barium sulfate was obtained.

Calculate the theoretical yield,_________________

Calculate the percentage yield._________________

Answer 1

Ba(NO3)2 (aq) + Na2SO4 (aq) ----------------> BaSO4 (s) + 2 NaNO3 (aq)

Given ,

amount of Ba(NO3)2 = 75 g and amount of Na2SO4 = 72 g

Since amount of both reactants are we have to find the limiting reactant which gives smaller amount of product by considering moles.

75 g Ba(NO3)2 = (75g / 261g/mol) = 0.287 moles

72g Na2SO4 = (72g /142g/mol) = 0.507 moles

Now,

BaSO4 obtained from 0.287 mol of Ba(NO3)2 = 0.287 mol [since 1 mol of Ba(NO3)2 gives 1 mol of BaSO4]

BaSO4 obtained from 0.507 mol of Na2SO4 = 0.507 mol [since 1 mol of Na2SO4 gives 1 mol of BaSO4]

Thus, Ba(NO3)2 is the limiting reactant.

Hence,

BaSO4 formed = 0.287 mol * 233.4 g/mol = 66.99 g = 67 g (approx)

So,

Theoretical yield = 67 g

And actual yield is given as 64.45 g

So,

Percentage yield = (actual yield / theoretical yield ) * 100

Percentage yield = (64.45 / 67) * 100 = 96.2 %

So,

Percentage yield = 96.2%