A 16.0 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.

Question

Nero

Chemistry

2 June, 02:56

A 16.0 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.

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Answers (1)

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Marlie Randall · Answer 1

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%

Answer:

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.