Limiting Reactant Numericals

Q1.   How many gram of NH₃ are formed when 100 g of each of the following reagents are reacted together according to following equation:                        

2NH₄Cl    +    Ca(OH)₂     →    CaCl₂    +    2NH₃   +   2H₂O

Solution

(I) Conversion of mass of each reactant into moles

Mass of NH₄Cl             = 100 g

Mass of Ca(OH)₂             = 100 g

Molar mass of NH₄Cl    = 14  +  4  +  35.5 =   53.5 g/mol

Molar mass of Ca(OH)₂= 40  + 32 +    2    =   74   g/mol

No. of moles (n) of NH₄Cl         = mass/molar mass = 100/53.5 = 1.869 mol

No. of moles (n) of Ca(OH)₂       = mass/molar mass = 100/74    = 1.351mol

(II) Calculation of molar amount of product from molar amount of each reactant

2NH₄Cl    +    Ca(OH)₂     →    CaCl₂    +    2NH₃   +   2H₂O

(Given)                                                        (Requreid)

2 mole             1 mole        --------------------> 2 mole

1.869 mole       1.315 mole -------------------> ?

            2 moles of NH₄Cl produces 2 mole NH₃

            1 mole   of NH₄Cl produces 2/2 mole NH₃

            1.869 mole of NH₄Cl produces 2/2 x 1.869 mole NH₃ =   1.869 mole NH₃

            1 mole   of Ca(OH)₂ produces 2 mole NH₃

            1.351 mole of Ca(OH)₂ produces 2/1 x 1.351 mole NH₃ =   2.7 mole NH₃

(III)       Determination of Limiting Reactant

From above calculation, it is clear that least amount of product (NH₃) is produced by NH₄Cl so NH₄Cl is a limiting reactant and amount of product is calculated from its amount.

(IV)      Conversion of Molar Amount of Product into Gram

Molar amount of NH₃= 1.869 moles

Molecular mass of NH₃ = 14 + 3  =  17 g/mol

Amount of NH₃ in g = moles    x    gram molecular mass =1.869    x    17 = 31.773 g NH₃

Q2.  How many grams of Na₂S₂O₃ will be produced when 200 g each of the three reagents are reacted together according to following equation?                        

 2Na₂S  +  Na₂CO₃  +  4SO₂  →  3Na₂S₂O₃  +  CO₂

Solution

1. Determination of molar amounts of reactants

Mass of each reactant = 200 g

Molar mass of Na₂S     = 46 + 32         = 78 g/mol

Molar mass of Na₂CO₃ = 46 + 12 + 48 = 106 g/mol

Molar mass of SO₂         = 32 + 32         = 64 g/mol

No. of moles of Na₂S in 200 g     = mass/molar mass = 200/78   =  2.56 moles

No. of moles of Na₂CO₃ in 200 g = mass/molar mass = 200/106 =  1.88 moles          

No. of moles of SO₂ in 200 g        = mass/molar mass = 200/64  =  3.12 moles

2. Calculation of molar amount of products from molar amount of each reactant

2Na₂S         +  Na₂CO₃  +  4SO₂            →  3Na₂S₂O₃  +  CO₂

(Given)          (Given)        (Given)                 (Required)

 2 mole           1 mole        4 mole ------------->  3 mole

 2.56 mol        1.88 mol     3.12 mol ---------->  ?

2 moles of Na₂S produces 3 moles Na₂S₂O₃

1 mole of Na₂S produces 3/2 moles Na₂S₂O₃

.56 moles of Na₂S produces 3/2 x 2.56 moles Na₂S₂O₃ = 3.84 moles Na₂S₂O₃

1 mole of Na₂CO₃ produces 3 moles Na₂S₂O₃

1.88 mole of Na₂CO₃ produces 3 x 1.88 moles Na₂S₂O₃ = 5.64 moles Na₂S₂O₃

4 moles of SO₂ produces 3 moles Na₂S₂O₃

1 mole of SO₂ produces ¾ moles Na₂S₂O₃

3.12 mole of SO₂ produces ¾ x 3.12 moles Na₂S₂O₃ = 2.34 moles Na₂S₂O₃

3.  Determination of L.R. and Mass of Product

From above calculation, it is clear that least amount of product (Na₂S₂O₃) is produced by SO₂, so SO₂ is a limiting reactants.  So amount of SO₂ is used to calculate amount of Na₂S₂O₃.

Molar amount of Na₂S₂O₃ = 2.34 moles

Molar Mass of  Na₂S₂O₃  =  2(23) + 2(32) + 3(16) = 46 + 64 + 48 = 158 g/mole

Amount of Na₂S₂O₃  in g = moles x molar mass

= 2.34 x 158

= 369.72 g Na₂S₂O₃  (Answer)