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 16.94

The volume of ammonia solution in (mL) is 35.62

Calculate the following:

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

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

Answer 1

lots of stuff going on here
1) CO2 + H2O ----> H2CO3
2) H2CO3 + NH3 ----> HCO3- + NH4+
3) Na+ + Cl- + HCO3- ----> NaHCO3 + Cl-
we could combine these into an overall equation
4) NaCl + H2O + CO2 + NH3 ----> NaHCO3 + NH4Cl

we have unlimited CO2 in 1 but only 0.160 mol of NH3 in 2
so we have a max of 0.160 mol HCO3 available
in 4 we have 0.251 mol NaCl and 0.160 mol NH3
ratio is 1 to 1 so NH3 LIMITS
giving 0.160 mol of NaHCO3
this is in 0.0399 L so we have 0.160 mol / 0.0399 L = 4.01 M
any over 0.75 M will precipitate
so 4.01 - 0.75 = 3.26 mol/L will precipitate
3.26 mol/L * 0.0399 L = 0.130 mol of NaHCO3 precipitates

theoretical yield is 0.160 mol * 84.02 g/mol = 13.4 g
precipitated is 0.130 mol * 84.02 g/mol = 10.9 g

finally when the ppt is recovered and dried by heat we get
2 NaHCO3 ----> Na2CO3 + H2O + CO2
the ratio is 2 to 1 so we get 0.065 mol of Na2CO3
0.065 mol * 106.01 g/mol = 6.89 g of Na2CO3