Question

22.2 g 33.3 g g g Limiting reactant:

Answer 1

PART 1 : ANSWER : MASS OF NbC = 26.23 g and MASS OF CO = 17.5 g

Consider a reaction : 14 C + 2 Nb₂O₅   4 NbC + 10 CO

From reaction , stoichiometric ratio of reactants is C : Nb₂O₅   = 14: 2= 7: 1

No of moles of C = mass / Molar mass

= 22.2 g / 12.01 g / mol

= 1.85 mol

Molar mass of Nb₂O₅ = ( 2 x 92.91 ) + ( 5 x 16.00 ) = 265.8 g / mol

No of moles of Nb₂O₅ = 33.3 g / 265.8 g / mol

= 0.125 mol

Provided molar ratio of reactants is C : Nb₂O₅   = 1.85 : 0.125 = 14.8 : 1

Comparing provided molar ratio with stoichiometric ratio , it is found that Nb₂O₅ is limiting reactant and yield of product depends upon mass of Nb₂O₅.

From reaction , 2 moles Nb₂O₅  $\equiv$ 4 mole NbC

Therefore, 0.125 moles Nb₂O₅  $\equiv$ 4 x 0.125 / 2  mole NbC

$\equiv$ 0.25 mole

Molar mass of NbC= 92.91 + 12.01 = 104.9 g / mol

0.25 mol NbC = 0.25 mol x ( 104.9 g / 1 mol )

= 26.23 g

From reaction , 2 moles Nb₂O₅  $\equiv$ 10 mole CO

Therefore, 0.125 moles Nb₂O₅  $\equiv$ 10 x 0.125 / 2  mole CO

$\equiv$ 0.625 mole CO

Molar mass of CO =12.01 + 16.00 = 28.01 g / mol

0.625 mol CO = 0.625 mol x ( 28.01 g / 1 mol )

= 17.5 g

PART 2 :

Consider reaction : 3 Cl ₂ + 6 NaOH 5 NaCl + NaClO ₃  + 3 H₂O

Moles of NaOH = Mass / Molar mass

= 15.0 g / 40.0 g/ mol

= 0.35 mol

From reaction , 6 moles NaOH  $\equiv$3 mole Cl ₂

Therefore, 0.35 moles NaoH  $\equiv$ 3 x 0.35 / 6 mole Cl ₂

$\equiv$ 0.175 mole Cl ₂

Molar mass of Cl ₂ = 2 x 35.45 = 70.9 g / mol

0.175 mole Cl ₂ = 0.175 mol x ( 70.9 g / 1 mol ) = 12.41 g

From reaction , 6 moles NaOH  $\equiv$5 mole NaCl

Therefore, 0.35 moles NaoH  $\equiv$ 5 x 0.35 / 6 mole NaCl

$\equiv$ 0.292 mole NaCl

Molar mass of NaCl = 22.99+ 35.45 = 58.44 g / mol

0.292 mole NaCl = 0.292 mol x ( 58.44 g / 1 mol ) =17.06 g

From reaction , 6 moles NaOH  $\equiv$1 mole NaClO₃

Therefore, 0.35 moles NaoH  $\equiv$ 1 x 0.35 / 6 mole NaClO₃

$\equiv$ 0.0583 mole NaClO₃

Molar mass of NaClO₃ = 22.99+ 35.45 + ( 3 x 16.00) = 106.44 g / mol

0.0583 mole NaClO₃ = 0.0583 mol x ( 106.44 g / 1 mol ) =6.20 g

From reaction , 6 moles NaOH  $\equiv$3 mole H₂O

Therefore, 0.35 moles NaoH  $\equiv$3 x 0.35 / 6 mole H₂O

$\equiv$0.175 mole H₂O

Molar mass of H₂O = ( 2 x 1.0079 ) + 16.00 = 18.02 g / mol

0.175 mole H₂O = 0.175 mol x ( 18.02 g / 1 mol ) = 3.15 g