Solvay PreLab A weighed amount of sodium chloride is completely dissolved in a m

Question

Solvay PreLab

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 util 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.

Data to be used for calculating the results
-The mass of sodium chloride in (g) is 14.5
-The volume of ammonia solution in (mL) is 23.6

Calculate the following
-How many moles of sodium chloride are present in the original solution?
-How many moles of ammonia are present originally?
-What is the limiting reagent? a)ammonia, b)sodium chloride
-What is the theoretical yield of sodium bicarbonate in grams?
-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 the precipitate?

Explanation / Answer

From the given data,

- moles of NaCl present in original solution = 14.5 g/58.4 g/mol = 0.25 mol

- moles NH3 present in original solution = 4.0 M x 0.0236 L = 0.0944 mol

- limiting reactant a) ammonia

as it is lower moles of the two reactant.

- theoretical yield of NaHCO3

moles NaHCO3 produced = 0.0944 mol

mass NaHCO3 = 0.0944 mol x 84.007 g/mol = 7.93 g

- mass of NaHCO3 expected = 7.93 g

- moles of Na2CO3 produced = 0.0944/2 = 0.0472 mol

mass of Na2CO3 to be produced = 0.0472 mol x 106 g/mol = 5.02 g

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