Answer:
Theoretical yield of potassium nitrate = 1.21 g
Theoretical yield of lead iodide = 2.77 g
Explanation:
Given data:
Mass of Lead nitrate = 2.00 g
Mass of potassium iodide = 3.00 g
Theoretical yield = ?
Percent yield = ?
Solution:
Chemical equation:
Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃
Now we will calculate the number of moles of lead nitrate.
Number of moles = mass/ molar mass
Number of moles = 2.00 g/ 331.2 g/mol
Number of moles = 0.006 mol
Number of moles of KI:
Number of moles = mass/ molar mass
Number of moles = 3.00 g/ 166 g/mol
Number of moles = 0.02 mol
Now we will compare the moles of lead nitrate with potassium iodide with lead iodide and potassium nitrate.
Pb(NO₃)₂ : PbI₂
1 : 1
0.006 : 0.006
Pb(NO₃)₂ : KNO₃
1 : 2
0.006 : 0.006×2 =0.012
KI : PbI₂
2 : 1
0.02 : 1/2×0.02 = 0.01
KI : KNO₃
2 : 2
0.02 : 0.02
Theoretical yield of both product depend upon lead nitrate because it is limiting reactant.
Theoretical yield of potassium nitrate:
Mass = number of moles × molar mass
Mass = 0.012 mol × 101.1 g/mol
Mass = 1.21 g
Theoretical yield of lead iodide:
Mass = number of moles × molar mass
Mass = 0.006 mol × 461.01 g/mol
Mass = 2.77 g
Percentage yield can not be determine because actual yield is not given.
