Chemistry, 01.03.2020 21:11 Hollywood0122

Theoretical yield and percentage yield of:

2.00g of Lead (II) Nitrate & 3.00g of Potassium Iodide

The balanced chemical equation is
Pb(NO3)2 + 2KI —> PbI2 + 2KO3

Please include the working out

Answers: 3

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Answer from: markrandsdenn

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.

Answer from: Quest

The answer for this question is d. none of these

Answer from: Quest

може ли да зареждате повече информация, моля

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Theoretical yield and percentage yield of:

2.00g of Lead (II) Nitrate & 3.00g of Pot...

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