A0.5674 g piece of copper is added to 10.00 ml; of 16 m hno3, producing 0.8024 g of copper (ii) nitrate. what is the precent yield of the reaction?

Percent yield = 0.075%

Explanation:

Given data:

Mass of copper = 0.5674 g

Volume of HNO₃ = 10 mL

Molarity of HNO₃  = 16 M

Experimental yield = 0.8024 g

Percent yield = ?

Solution:

First of all we will write the balance chemical equation:

 4HNO₃  + Cu  → Cu(NO₃)₂ + 2NO₂ + 2H₂O

First of all we will calculate the number of moles of  HNO₃.

Number of moles = Molarity × volume

Number of moles = 16 mol/L  × 0.01 L

Number of moles =  0.16 mol

Mass of HNO₃ = moles × molar mass

Mass of HNO₃ =  0.16 mol × 63.01 g/mol

Mass of HNO₃ = 10.0816 g

Now we will calculate the moles of copper.

Number of moles = mass/molar moles

Number of moles = 0.5674 / 63.546

Number of moles = 0.009 mol

Now we will compare the moles of copper nitrate with copper and HNO₃.

            HNO₃     :     Cu(NO₃)₂

              4          :        1

               0.16     :      1/4×0.16 = 0.04 mol

             Cu        :      Cu(NO₃)₂

              1           :         1

         0.009     ;       0.009

The number of moles produce by copper are less so copper will be limiting reactant.

 Mass of copper nitrate = moles  ×  molar mass

 Mass of copper nitrate = 0.009  mol×  187.56 g/mol

  Mass of copper nitrate = 1.68 g

Percent yield:

Percent yield = (actual yield / theoretical yield ) × 100

Percent yield =  (0.8024 g/ 1.68 g) × 100

Percent yield = 0.075%

it's all four options.

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