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 4 mole NbC
Therefore, 0.125 moles Nb2O5 4 x 0.125 / 2 mole NbC
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 10 mole CO
Therefore, 0.125 moles Nb2O5 10 x 0.125 / 2 mole CO
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 3 mole Cl 2
Therefore, 0.35 moles NaoH 3 x 0.35 / 6 mole Cl 2
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 5 mole NaCl
Therefore, 0.35 moles NaoH 5 x 0.35 / 6 mole NaCl
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 1 mole NaClO3
Therefore, 0.35 moles NaoH 1 x 0.35 / 6 mole NaClO3
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 3 mole H2O
Therefore, 0.35 moles NaoH 3 x 0.35 / 6 mole H2O
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
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