Stoichiometry numerical

## Stoichiometry numerical-

**Question 1) How many moles of N2 are needed to produce 8.2 moles of NH3 by the reaction of N2 & H2 ?**

**Solution –**

N2(g) + 3H2(g) –> 2NH3(g)

1 mole 3 mole 2 mole

2 mole NH3 is produced from = 1 mole N2

8.2 mole NH3 is produced from = 1x 8.2 / 2 mole N2= 4.1 mole N2

**Moles of N2 are needed = 4.1 mole**

**Question 2) What weight of Zinc would be required to produce enough H2 to reduce completely 10.2 gm Copper (II) oxide to Copper?**

**Solution –**

(a) CuO + H2 —-> Cu + H2O

63.5+16= 79.5 gm 1×2= 2gm

79.5 gm CuO requires H2 for reduction = 2 gm

10.2 gm CuO requires H2 for reduction = 2x 10.2 / 79.5 gm= 0.257 gm

mass of H2 required for reduction of CuO to Cu = 0.257 gm

(b) Zn + H2SO4 —–> ZnSO4 + H2

65 gm 2 gm

2 gm H2 is produced from = 65 gm zinc

0.257 gm H2 is produced from = 65 x 0.257 /2.0 gm zinc = 8.3525 gm

**Weight of Zinc required to produce 0.257 gm H2 = 8.35 gm **

**Question 3) Calculate the amount of KClO3 needed to supply sufficient oxygen for burning 112 L of CO gas at NTP.**

**Solution –**

CO (g) + 1/2 O2(g) —–> CO2(g)

1 mole 1/2 = 0.5 mole

Volume of 1 mole CO at NTP = 22.4L

22.4 Litre of CO at NTP requires = 0.5 mole O2

112 Litre of CO at NTP requires = 0.5 x 112 / 22.4 mole O2

Moles of O2 = 2.5

2KClO3. —> 2KCl + 3O2

2 mole 3 mole

3 moles O2 is produced from = 2 moles KClO3

2.5 moles O2 is produced from = 2 x 2.5 / 3.0 moles KClO3

Moles of KClO3 required = 1.67

molar mass of KClO3 = 39 + 35.5 + 16×3= 122.5 gm/mole

mass of KClO3 = mole x molar mass = 1.67 x 122.5 = 204.575 gm

**Mass of KClO3 needed = 204.58 gm**

**Stoichiometry numerical-**

**Question 4) 50.0 Kg N2 & 10.0 Kg of H2 are mixed together to produce NH3.Calculate the amount of NH3 produced in gm. Identify the limiting reagent.**

**Solution-**

N2(g) + 3H2 (g) —-> 2NH3(g)

mole ratio 1 mole 3 mole 2 mole

molar mass of N2 = 14×2 = 28 gm/mole

molar mass of H2 = 1×2 = 2 gm/mole

molar mass of NH3 = 14+1×3 = 17 gm/mole

mass of N2 = 50.0 Kg = 50.0 x 1000 gm = 50000 gm

Moles of N2 = mass in gm/molar mass

= 50000/28

**Moles of N2= 1.786 x 10^{3}**

mass of H2 = 10.0 Kg = 10.0 x 1000 gm = 10000 gm

Moles of H2 = mass in gm/molar mass

= 10000/2

Moles of H2 = 5.0 x *10*^{3}

1 mole N2 reacts with = 3 mole H2

1.786 x *10*^{3} mole N2 reacts with =1.786 x *10*^{3} x 3 mole H2

Moles of H2 required = 5.358 x *10*^{3}

But we have only 5.0 x *10*^{3} Moles of H2, so **H2 is limiting reagent.**

3 moles H2 on reaction gives= 2 moles NH3

5.0 x *10*^{3} moles H2 on reaction gives= 2 x 5.0 x *10*^{3} / 3 moles NH3

Moles of NH3 = 3.33 x *10*^{3}

mass of NH3 = moles x molar mass = 3.33 x *10*^{3} x 17 = 56.61 x 1*0*^{3}

**Mass of NH3 = 56.61 x 10^{3} gm = 56610 gm**

**Mass of NH3 =56610 gm**