If you mix a 25.0 mL sample of a 1.20 M potassium phosphate solution with 30.0 m

Question

If you mix a 25.0 mL sample of a 1.20 M potassium phosphate solution with 30.0 mL of a 1.10 M barium nitrate solution, a precipitation reaction occurs. You collect 3.31 grams of dry solid precipitate. What is the % yield in your reaction? 94.8% 84.5% 36.7% 50.0% 66.7% Which of the following is a weak electrolyte in aqueous solution? ammonium carbonate hydrofluoric acid CH_3CH_2OH CO_2 HClO_4 In each case two aqueous solutions are mixed. In which one will No precipitate form? Lithium carbonate and lead(II) nitrate Copper(II) nitrate and calcium sulfide Ammonium sulfate and strontium chloride Iron (II) acetate and sodium phosphate Nickel (II) chloride and potassium sulfate Aqueous solutions containing each pair of compounds given are mixed. Which gives an incorrect net ionic equation for the reaction that occurs? Sodium phosphate and nickel(II) chloride 3 Ni^2+ (aq) + 2 PO_4^3- (aq) rightarrow Ni_3(PO_4)_2(S) Potassium hydroxide and ammonium sulfate 2 OH^- (aq) + 2 NH4 (aq) rightarrow 2 NH_3(g) +2H_2O(I) Sodium carbonate and calcium nitrate CO_3^2_ (aq) + Ca^2+(aq) rightarrow CaCO_3(S) Lithium sulfite and hydrochloric acid SO_3^2- (aq) + 2 H^+ (aq) rightarrow SO_2(g) + H_2O(I) Magnesium chloride and copper(II) sulfate Cu^2+ + 2 Cl^- (aq) rightarrow CuCl_2(s)

Explanation / Answer

3 Ba(NO3)2 + 2K3PO4 --> 6KNO3 + Ba3(PO4)2

First calculate the moles of K3PO4 = Molarity * volume in L

= 1.20 M * 0.025L

= 0.03 moles K3PO4

the moles of Ba(NO3)2 = Molarity * volume in L

= 1.10*0.030

= 0.033 moles Ba(NO3)2

Here Ba(NO3)2 is limiting agent due to following reasons:

Now calculate the moles of Ba3(PO4)2 :

0.033 moles Ba(NO3)2 *1 mole Ba3(PO4)2 /3 moles Ba(NO3)2

= 0.011 mole Ba3(PO4)2

Amount of Ba3(PO4)2 = number of moles * molar mass

= 0.011 mole Ba3(PO4)2 *601.93 g/mol

=6.62 g Ba3(PO4)2

% yield Ba3(PO4)2 = observed yield / theoretical yield *100

= 3.31 g /6.62 g*100

= 50.0 %

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