s-Block ElementsMedium
Question
One g equimolecular mixture of Na2CO3 and NaHCO3 is reacted with 0.1N HCl. The mL of 0.1 N HCl required to react completely with the above mixiture is :
Options
A.15.78mL
B.153.8mL
C.198.4mL
D.295.5mL
Solution
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
NaHCO3HCl → NaCl + H2O + CO2
Mole of Na2CO3 =
0.0047
Mole of NaHCO3 =
= 0.0059
0.0047 mole of Na2CO3 ≡ 0.0047 × 2 mole of HCl
= 0.0094 mole of HCl 0.0059mole of 3 NaHCO3 ≡ 0.0059mole of HCl
Total mole required = 0.0059 moles of HCL If V be the volunes in litre of HCl required them mole of HCl required = V × 0.1
V =
= 0.153 litre
NaHCO3HCl → NaCl + H2O + CO2
Mole of Na2CO3 =
0.0047Mole of NaHCO3 =
= 0.00590.0047 mole of Na2CO3 ≡ 0.0047 × 2 mole of HCl
= 0.0094 mole of HCl 0.0059mole of 3 NaHCO3 ≡ 0.0059mole of HCl
Total mole required = 0.0059 moles of HCL If V be the volunes in litre of HCl required them mole of HCl required = V × 0.1
V =
= 0.153 litreCreate a free account to view solution
View Solution FreeMore s-Block Elements Questions
Which of the following disproportionate(s) on heating with sodium hydroxide ?...Which of the following statement is incorrect ?...In the manufacturing of sodium carbonate (Na2CO3), using Solvay process, which of the following compound is used ?...The first ionization potential is maximum for :...A chloride dissolves appreciably in cold water. When placed on a platinum wire in Bunsen flame, no distinctive colour is...