Calculate the percent yield (%)if11.083gof BaSO4precipitates when1.350Lof 0.0550 M BaI2 and1.200 Lof0.0476M?

Question

For the reaction BaI2 + MgSO4 (aq)BaSO4 precipitates out.Calculate the percent yield (%)if11.083gof BaSO4precipitates when1.350Lof 0.0550 M BaI2 and1.200 Lof0.0476 M MgSO4(aq)are mixed.

Answer 1

First find how many moles of each reactant you have

moles BaI2 = M*V= 1.35*0.055
moles MgSO4 = M*V= 1.2*0.0476

The reaction is
BaI2 + MgSO4 ->BaSO4 + MgI2

the stoichiometry is 1:1, so you should have equal moles of the two reactants. In your case however one of them is in excess. So you will have to calculate the yield based on the limiting reagent.

Since the stoichiometry is also 1:1 for BaSO4 and the limiting reagent you should have moles of BaSO4 = moles of the limiting reactant. But moles=mass/MW => mass= moles*MW
You have calculated the moles, find the MW of BaSO4 and you'll get the theoretical mass of BaSO4 that would come from 100% reaction.
The percent yield is (actual mass/theoretical mass)*100%

Answer 2

BaCl2 + MgSO4 ------> BaSO4 + MgCl2
1.350 L of 0.55 M BaCl2 contains 0.7425 moles of BaCl2.
1.2 L of 0.476 M MgSO4 contains 0.5712 moles of MgSO4.
since MgSO4 is the limiting reactant, it controls the reaction.
so theoretically, only 0.5712 moles of BaSO4 are possible.
this represents 100% yield.
you got yield of 11.083 g of BaSO4. which contains
[11.083/Mwt of BaSO4] =11.083/233.4 = 0.0472moles of BaSO4
so your percent yield is [0.0472/0.5712] * 100 = 8.26%
:)

Answer 3

i got 9.98% but im not a chemist. i took 0.0476 times the molecular weight of BaSO4 (233.33) and got a theoretical yield of 111.067 g and divied 11..083 by that and got 9.98%