A 17.0 mL sample of a 1.62 M potassium sulfate solution is mixed with 14.8 mL of

Question

A 17.0 mL sample of a 1.62 M potassium sulfate solution is mixed with 14.8 mL of a 0.890 M barium nitrate solution and this precipitation reaction occurs:

K2SO4(aq)+Ba(NO3)2(aq)BaSO4(s)+2KNO3(aq)

The solid BaSO4 is collected, dried, and found to have a mass of 2.48 g . Determine the limiting reactant, the theoretical yield, and the percent yield.

Part A

Determine the limiting reactant.

Express your answer as a chemical formula.

K2SO4+Ba(NO3)2BaSO4+2KNO3

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Part B

Determine the theoretical yield.

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Part C

Determine the percent yield.

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K2SO4+Ba(NO3)2BaSO4+2KNO3

Explanation / Answer

A)

volume of K2SO4, V = 17.0 mL
= 1.7*10^-2 L


we have below equation to be used:
number of mol in K2SO4,
n = Molarity * Volume
= 1.62*0.017
= 2.754*10^-2 mol
volume of Ba(NO3)2, V = 14.8 mL
= 1.48*10^-2 L


we have below equation to be used:
number of mol in Ba(NO3)2,
n = Molarity * Volume
= 0.89*0.0148
= 1.317*10^-2 mol
we have the Balanced chemical equation as:
K2SO4 + Ba(NO3)2 ---> BaSO4 + 2 KNO3


1 mol of K2SO4 reacts with 1 mol of Ba(NO3)2
for 2.754*10^-2 mol of K2SO4, 2.754*10^-2 mol of Ba(NO3)2 is required
But we have 1.317*10^-2 mol of Ba(NO3)2

so, Ba(NO3)2 is limiting reagent

B)
we will use Ba(NO3)2 in further calculation


Molar mass of BaSO4 = 1*MM(Ba) + 1*MM(S) + 4*MM(O)
= 1*137.3 + 1*32.07 + 4*16.0
= 233.37 g/mol

From balanced chemical reaction, we see that
when 1 mol of Ba(NO3)2 reacts, 1 mol of BaSO4 is formed
mol of BaSO4 formed = (1/1)* moles of Ba(NO3)2
= (1/1)*1.317*10^-2
= 1.317*10^-2 mol


we have below equation to be used:
mass of BaSO4 = number of mol * molar mass
= 1.317*10^-2*2.334*10^2
= 3.07 g
Answer: 3.07 g

C)

% yield = actual mass*100/theoretical mass
= 2.48*100/3.074
= 80.7 %
Answer: 80.7 %

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