How many mL of 0.1 M HCL are required to react completely with 1g mixture of Na2CO3 and NaHCO3 containing equimolar amount of both?
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Answer:
158 mL
Total mol of Hcl required to react with a mixture of NaHCO3 & Na2Co3= 2x 0.00526 + 0.00526= 0.01578 mol. Hence, 158 mL of 0.1 M of HCl is required to react completely with 1 g mixture of Na₂Co₃ and NaHCO₃, containing equimolar amounts of both.
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