question_answer 18)

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and  containing equimolar amounts of the two?

Answer:

  2NaCl + CO2 + H2O             NaCl + CO2 + H2O   Mass of a mixture containing 1 mol each of Na2CO3 and NaHCO3 = (106 + 84) g = 190 g 190 g of mixture requires HCl = 3 mol 1 g of mixture requires HCl   Moles of solute (HCl) = M × V(L)