Question

Theoretical yield and percentage yield of:

Answer 1

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.