Question

Please go step-by-step.

A 24.0 ml sample of 1.16M potassium sulfate is mixed with 14.3 ml of a 0.880 barium nitrate solute this precipitation reaction occurs: K_2 SO_4+Ba(NO_3 )_2→BaSO_4+2KNO_3

Determine the limiting reactant

Determine the theoretical yield

The solid BaSO_4 is collected, dried, and found to have a mass of 2.53 g. Calculate the percent yield

What is/are the concentration/s of ions in solution after the reaction occurs?

Answer 1

K2SO4 + Ba(NO3)2 --------------> BaSO4 + 2KNO3

moles of K2SO4 = 1.16 x 24/1000 = 0.0278

moles of Ba(NO3)2 = 0.88 x 14.3 / 1000 = 0.0126

according to balanced reaction

1 mole K2SO4 reacts with 1 mole Ba(NO3)2

0.0278 moles K2SO4 reacts with 0.0278 x 1 / 1 = 0.0278

but we have only 0.0126 moles Ba(NO3)2.

Limiting reagent = Ba(NO3)2

Ba(NO3)2 and BaSO4 are 1 : 1 ratio.

so therotical yield = 0.0126 moles

mass = 0.0126 x 233.38 = 2.94 g

therotical yield = 2.94 g

actual yield = 2.53 g

% yield = (2.53 / 2.94) x 100

% yield = 86.05