To a 150 mL beaker were added 0.510 g of solid copper powder and 2.5 mL of a 15

Question

To a 150 mL beaker were added 0.510 g of solid copper powder and 2.5 mL of a 15 M aqueous solution of nitric acid. Brown fumes were evolved, and 25 mL of water was then added to the blue solution formed. To the aqueous solution was added 3.010 g of solid sodium carbonate in small portions. Only after all of the excess hydrogen ion in the aqueous solution was consumed by reaction with added carbonate ion did carbonate ion precipitate copper (II) ion as copper (II) carbonate. The solid copper (II) carbonate was isolated by vacuum filtration and dried; 0.920 g of copper (II) carbonate was collected. The net ionic equations for the reactions which occurred are shown below. Cu(s) + 4H^+(aq) + 2NO_3(aq) rightarrow Cu^2+(aq) + 2NO_2(g) + 2H_2O(l) 2H^+(aq) + CO^2-_3 (aq) rightarrow H_2O(l) + CO_2(g) Cu^2+(aq) + CO^2-_3 (aq) rightarrow CuCO_3(s) Calculate the percent yield of copper (II) carbonate.

Explanation / Answer

Solution :-

Cu^2+ + CO3^2-   ----- > CuCO3(s)

Mole ratio of the Cu2+ and CO3^2- is 1 : 1

So lets calculate the theoretical yield of the product using the mass of the Cu used in the reaction

0.510 g Cu * 1 mol / 63.546 g = 0.008026 mol Cu2+

Since mole ratio is 1 : 1

So the moles of the CuCO3 that can be produced = 0.008026 moles

Now lets convert this moles of copper carbonate to its mass

Mass = moles * molar mass

Mass of CuCO3 = 0.008026 mol * 123.554 g per mol

                            = 0.9916 g CuCO3

Now lets calculate the percen yield of the reaction

% yield = (actual yield / theoretical yield )*100%

               = (0.920 g / 0.9916 g)*100%

               = 92.78 %

So the percent yield of the reaction is 92.78 %

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