Answer:
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%
