Question

The mass of sodium chloride in (g) is 14.19 The volume of ammonia solution in (mL) is 36.15 Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer 1

This is an incomplete question, here is a complete 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 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 14.19

-The volume of ammonia solution in (mL) is 36.15

Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer : The theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Explanation :

First we have to calculate the moles of NaCl and
`NH_3`.

`\text{ Moles of }NaCl=\frac{\text{ Mass of }NaCl}{\text{ Molar mass of }NaCl}=(14.19g)/(58.5g/mole)=0.243moles`

`\text{ Moles of }NH_3=\text{ Concentration of }NH_3* \text{ Volume of solution}=4.00M* 0.3615L=1.446moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

`NH_3+NaCl+CO_2+H_2O\rightarrow NaHCO_3+NH_4Cl`

From the balanced reaction we conclude that

As, 1 mole of
`NaCl` react with 1 mole of
`NH_3`

So, 0.243 mole of
`NaCl` react with 0.243 mole of
`NH_3`

From this we conclude that,
`NH_3` is an excess reagent because the given moles are greater than the required moles and
`NaCl` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`NaHCO_3`

From the reaction, we conclude that

As, 1 mole of
`NaCl` react to give 1 mole of
`NaHCO_3`

So, 0.243 moles of
`NaCl` react to give 0.243 moles of
`NaHCO_3`

Now we have to calculate the mass of
`NaHCO_3`

`\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3* \text{ Molar mass of }NaHCO_3`

Molar mass of sodium bicarbonate = 84 g/mol

`\text{ Mass of }NaHCO_3=(0.243moles)* (84g/mole)=20.4g`

Thus, the theoretical yield of sodium bicarbonate in grams is, 20.4 grams.