A16.0 ml sample of a 1.04 m potassium sulfate solution is mixed with 14.3 ml of a 0.880 m barium nitrate solution and this precipitation reaction occurs: k 2 s o 4 (aq)+ba(n o 3 ) 2 (aq)→bas o 4 (s)+2kn o 3 (aq) the solid bas o 4 is collected, dried, and found to have a mass of 2.60 g . determine the limiting reactant, the theoretical yield, and the percent yield.

An insoluble salt is created when solutions having soluble salts are reacted. The salt formed in the reaction which gets separated out is called insoluble salt.

The limiting agent is Barium Nitrate and theoretical and the percentage yields are:2.94 g and 88.5%.

This can be explained as:

K₂SO₄(aq) + Ba(NO₃)₂ (aq)  → BaSO₄ (s) + 2 KNO ₃(aq)

Number of moles in the K₂SO₄ sample

Number of moles in the Ba(NO₃)₂ sample

The ratio between the two reactants is 1:1 so the limiting factor will be the one that is present in the less quantity. Hence, Ba(NO₃)₂ Barium Nitrate is the limiting factor.

The molecular mass of BaSO₄ =  

The theoretical yield of Barium Sulphate is

Actual yield = 2.60 g

The percentage yield

Therefore, the limiting factor is Barium Nitrate.

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Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol

Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol

Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.

The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%

 
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%

I apologize for the mistake previous to this update.

Ineed with this also i hope you understand it

It could possibly be beneficial if the experiment is successful.