Question

Sodium chlorate, NaClO3, can be prepared by the following series of reactions -
1) 2KMnO4 + 16HCl ——> 2KCl + 2 MnCl2 + 8H2O + 5Cl2

2) 6cl2 + 6Ca(OH)2 –> Ca(ClO3)2 + 5CaCl2 + 6H2O

3) Ca(ClO3)2 + Na2SO4 –> CaSO4 + 2NaClO3

What mass percent of NaCl O3 can be prepared from hundred CC of concentrated HCL ( density =1.18 g/cm³, 36 mass % of HCl )? Assume alert the other substances are present in excess amounts.

Answer 1

Supposing the first given equation to actually be:

... → 2 MnCl2 + 2 KCl ...

Supposing the second given equation to actually be:

6 Cl2 + 6 Ca(OH)2 ...

Multiply the first given equation by 6, the second by 5 and the third by 5:

12 KMnO4 + 96 HCl → 12 MnCl2 + 12 KCl + 48 H2O + 30 Cl2

30 Cl2 + 30 Ca(OH)2 → 5 Ca(ClO3)2 + 25 CaCl2 + 30 H2O

5 Ca(ClO3)2 + 5 Na2SO4 → 5 CaSO4 + 10 NaClO3

Add these three equations together algebraically:

12 KMnO4 + 96 HCl + 30 Cl2 + 30 Ca(OH)2 + 5 Ca(ClO3)2 + 5 Na2SO4 → 

12 MnCl2 + 12 KCl + 48 H2O + 30 Cl2 + 5 Ca(ClO3)2 + 25 CaCl2 + 30 H2O + 

5 CaSO4 + 10 NaClO3

Cancel like amounts of the same species from both sides:

12 KMnO4 + 96 HCl+ 30 Ca(OH)2 + 5 Na2SO4 → 12 MnCl2 + 12 KCl + 48 H2O + 

25 CaCl2 + 30 H2O + 5 CaSO4 + 10 NaClO3

Add the amounts of H2O on the right:

12 KMnO4 + 96 HCl+ 30 Ca(OH)2 + 5 Na2SO4 → 

12 MnCl2 + 12 KCl + 78 H2O + 25 CaCl2 + 5 CaSO4 + 10 NaClO3

What remains is the overall reaction.

(100 cc) x (1.18 g/cc) x (0.36) / (36.4611 g HCl/mol) x (10/96) x 

(106.4413 g NaClO3/mol) = 12.9 g NaClO3

Answer 2

HOLAA MATE

Answer:

Let the first given equation is - ... → 2 MnCl2 + 2 KCl ...

Let the second given equation is -  6 Cl2 + 6 Ca(OH)2

Multiply the first given equation by 6, second by 5 and third by 5

Thus,

12 KMnO4 + 96 HCl → 12 MnCl2 + 12 KCl + 48 H2O + 30 Cl2  

30 Cl2 + 30 Ca(OH)2 → 5 Ca(ClO3)2 + 25 CaCl2 + 30 H2O  

5 Ca(ClO3)2 + 5 Na2SO4 → 5 CaSO4 + 10 NaClO3  

Add the three equations together -  

12 KMnO4 + 96 HCl + 30 Cl2 + 30 Ca(OH)2 + 5 Ca(ClO3)2 + 5 Na2SO4 →  12 MnCl2 + 12 KCl + 48 H2O + 30 Cl2 + 5 Ca(ClO3)2 + 25 CaCl2 + 30 H2O +  5 CaSO4 + 10 NaClO3

Cancelling the like amounts of the same from both the sides:  

12 KMnO4 + 96 HCl+ 30 Ca(OH)2 + 5 Na2SO4 → 12 MnCl2 + 12 KCl + 48 H2O +  25 CaCl2 + 30 H2O + 5 CaSO4 + 10 NaClO3  

Adding the amounts of H2O on the right:  

12 KMnO4 + 96 HCl+ 30 Ca(OH)2 + 5 Na2SO4 →  

12 MnCl2 + 12 KCl + 78 H2O + 25 CaCl2 + 5 CaSO4 + 10 NaClO3  

The remaining overall reaction -

(100 cc) x (1.18 g/cc) x (0.36) / (36.4611 g HCl/mol) x (10/96) x  

(106.4413 g NaClO3/mol)

= 12.9 g NaClO3