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 15.61 The volume of ammonia solution in (mL) is 36.62

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?

Answer 1

Answer

1) Moles of NaCl = 15.61/58.44 = 0.2671

2) Moles of NH3 = 4*(36.62/1000) = 0.1465

3) Theoretical yield is 100% of Na+ ions. That is, 0.2671 moles.

4) Solubility of bicarbonate in 36.62 mL = 0.75*(36.62/1000) = 0.0275 moles.

Amount that will precipitate from this solution = 0.2671 - 0.0275 = 0.2396 moles or (moles*molar mass) 20.1281 g

5) Conversion of bicarbonate to carbonate is a equimolar reaction. (NaHCO3 + NaOH ---> Na2CO3 + H2O) Therefore, 0.2396 moles of Na2CO3 is equivalent 25.3976 g