Question

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed until the limiting reagent is entirely used up. The solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter.

Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate

The mass of sodium chloride in (g) is 11.09

The volume of ammonia solution in (mL) is 35.03

Calculate the following

(1) How many moles of sodium chloride are present in the original solution?

(2) How many moles of ammonia are present originally?

(3) What is the theoretical yield of sodium bicarbonate in grams?

(4) How many grams of the sodium bicarbonate formed will precipitate from the above volume of solution?

(5) How many grams of sodium carbonate will be produced from this precipitate?

This has to do with the Solvay Process

Answer 1

NaCl + CO2 + NH3 +H2O------> NaHCO3 + NH4Cl

no of moles of Nacl = W/G.M.Wt

                           = 11.09/58.5 = 0.189moles

no of moles of NH3 = Molarity * volume

                           = 4*0.0353 =0. 1412 moles

1 mole of Nacl react with 1 moles of NH3 s. NH3 is limiting reagent

1 mole of NH3 react with NaCl and CO2 to form 1 mole of NaHCO3

0.1412 moles of NH3 react with NaCl and Co2 to form 0.1412 moles of NaHCO3

mass of NaCO3 = no of moles * Gram molar mass

                        = 0.1412*84 = 11.86gm