Sodium chlorate, NaClO 3 , can be prepared by the following series of reactions?
2KMnO4 + 16 HCl→ 2MnCl +2KCl+ 8H2O + 5Cl2 then
6Cl+ 6Ca(OH)2→Ca(ClO3)2+ 5CaCl2 + 6H2O
Ca(ClO3)2+ Na2SO4→CaSO4+ 2NaClO3
What mass of NaClO3 can be prepared from 100 cc of concentrated HCl (density =1.18 g cm–3
, 36 mass % HCl)? Assume all other substances are present in excessamounts
Answers
Answer:
12.9 g NaClO3
Explanation:
Suppose the first given equation is - ... → 2 MnCl2 + 2 KCl ...
Suppose 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
Answer:
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Explanation:
