For the reaction BaI2 + MgSO4 (aq) BaSO4 precipitates out.?

Question

Calculate the percent yield (%) if 18.862 g of BaSO4 precipitates
when 2.150 L of 0.0600 M BaI2 and 1.980 L of 0.0501 M MgSO4 (aq) are mixed.

Answer 1

% yield is equal to the actual yield divided by the theoretical yield.

The balanced equation of the reaction is:
BaI2 + MgSO4 ---> BaSO4 + MgI2

From the information in the problem, we have:
2.150(0.0600) = 0.129 moles of BaI2
and
1.980(0.0501) = 0.0992 moles of MgSO4.

Therefore, your limiting reagent is MgSO4.

Because MgSO4 and BaSO4 have the same coefficients in the balanced equation, 0.0992 moles of BaSO4 should be produced by the reaction.

Your theoretical yield can be calculated by:
(molar mass of BaSO4: 233.37 g)
0.0992 mol BaSO4 x (233.37 g/ 1 mol BaSO4) = 23.2 g

So, % yield = 18.862/23.2 x 100% = 81.3%

(Note that sig figs will have a slight impact on calculations)

Hope that helped.

Answer 2

Find the moles of BaI2 and MgSO4:

n1 = C1*V1 = 0.06*2.15 = 0.129 mol BaI2

n2 = C2*V2 = 0.0501*1.980 = 00.992 mol MgSO4 (approx)

According to the balanced chem. equation:

BaI2 + MgSO4 --> BaSO4 + MgI2

the MgSO4 is the limiting reagent, so theoretically we get 0.0992 mol of BaSO4, or:

m = n*Mr = 0.0992*233.4 = 23.152 g of BaSO4, where Mr = 233.4 is the molar mass of BaSO4.

So the percent yield equals:

18.862*100/23.152 = 81.47%