a 0.5674 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 dа ta:

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%