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 14.99

The volume of ammonia solution in (mL) is 36.03

Calculate the following:

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

Well, let's calculate it step by step, using a bit of clown logic!

First, we need to find moles of sodium chloride (NaCl) present:

Moles of NaCl = mass / molar mass
Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol

So, moles of NaCl = 14.99 g / 58.44 g/mol

Now, let's stoichiometrically convert moles of NaCl to moles of sodium bicarbonate (NaHCO3):

1 mole of NaCl reacts with 1 mole of NaHCO3

Moles of NaHCO3 = moles of NaCl

Next, we'll convert moles of sodium bicarbonate to grams:

Mass of NaHCO3 = moles of NaHCO3 * molar mass of NaHCO3
Molar mass of NaHCO3 = 22.99 + 1.01 + 12.01 + 3 * 16.00 = 84.01 g/mol

Mass of NaHCO3 = moles of NaHCO3 * 84.01 g/mol

Finally, let's calculate the volume of solution in liters:

Volume of ammonia solution = 36.03 mL / 1000 mL/L

And we know that the solubility of sodium bicarbonate is 0.75 mol/L at ice temperature.

Now, we have all the pieces to solve the problem and calculate the grams of sodium bicarbonate precipitated:

Grams of NaHCO3 = Mass of NaHCO3 = (moles of NaCl) * (molar mass of NaHCO3)
= (14.99 g / 58.44 g/mol) * (84.01 g/mol)

So, in conclusion, the grams of sodium bicarbonate formed that will precipitate from the given volume of solution is the calculated value above.

And remember, if you don't find it funny, just pretend I'm wearing a clown wig to make it better!

To determine the grams of sodium bicarbonate formed, we first need to calculate the number of moles of sodium bicarbonate formed using the given information.

1. Calculate the number of moles of sodium chloride:
Given mass of sodium chloride = 14.99 g
Molar mass of sodium chloride (NaCl) = 58.44 g/mol

Number of moles of sodium chloride = mass / molar mass
Number of moles of sodium chloride = 14.99 g / 58.44 g/mol

2. Calculate the number of moles of sodium bicarbonate formed:
The balanced chemical equation representing the reaction is:
2 NH3(aq) + CO2(g) + NaCl(aq) → NaHCO3(s) + NH4Cl(aq)

Looking at the stoichiometry of the equation, we can see that 1 mole of sodium chloride reacts to form 1 mole of sodium bicarbonate.
Therefore, the number of moles of sodium bicarbonate formed will be equal to the number of moles of sodium chloride.

Number of moles of sodium bicarbonate formed = Number of moles of sodium chloride

3. Calculate the mass of sodium bicarbonate formed:
Given solubility of sodium bicarbonate in water at ice temperature = 0.75 mol/L
Volume of ammonia solution = 36.03 mL = 0.03603 L

The solubility value gives the maximum amount of sodium bicarbonate that can dissolve in 1 liter of water. Since the volume of ammonia solution used is less than 1 liter, the entire amount of sodium bicarbonate will precipitate out.

Mass of sodium bicarbonate formed = Number of moles of sodium bicarbonate formed × molar mass of sodium bicarbonate
Mass of sodium bicarbonate formed = Number of moles of sodium chloride × molar mass of sodium bicarbonate

Now, let's substitute the known values into the equation:
Mass of sodium bicarbonate formed = (14.99 g / 58.44 g/mol) × (84.01 g/mol)

Calculate the result using the above equation to find the grams of sodium bicarbonate formed.

To find the mass of sodium bicarbonate formed, we need to first calculate the moles of sodium bicarbonate that will precipitate from the given volume of solution.

Step 1: Convert the volume of ammonia solution from mL to liters.
Given: Volume of ammonia solution = 36.03 mL
1 L = 1000 mL
Therefore, Volume of ammonia solution = 36.03 mL / 1000 = 0.03603 L

Step 2: Calculate the moles of sodium chloride (NaCl) used.
Given: Mass of sodium chloride = 14.99 g
The molar mass of sodium chloride (NaCl) = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Moles of sodium chloride = Mass of sodium chloride / Molar mass of sodium chloride = 14.99 g / 58.44 g/mol

Step 3: Since sodium chloride is the limiting reagent, it reacts with ammonia and carbon dioxide to form sodium bicarbonate (NaHCO3).
The balanced equation is: 2 NH3 + CO2 + NaCl → NaHCO3 + 2 NH4Cl

Step 4: Determine the moles of sodium bicarbonate formed.
According to the balanced equation, 1 mole of sodium bicarbonate is formed for every mole of sodium chloride used.
So, Moles of sodium bicarbonate = Moles of sodium chloride

Step 5: Convert the moles of sodium bicarbonate to mass.
Given: Solubility of sodium bicarbonate in water = 0.75 mol/L
Therefore, Moles of sodium bicarbonate per liter = 0.75 mol/L

Mass of sodium bicarbonate formed = Moles of sodium bicarbonate * Molar mass of sodium bicarbonate

To summarize the calculations:

1. Convert the volume of ammonia solution to liters:
Volume of ammonia solution = 36.03 mL = 0.03603 L

2. Calculate the moles of sodium chloride:
Moles of sodium chloride = 14.99 g / 58.44 g/mol

3. Calculate the moles of sodium bicarbonate formed:
Moles of sodium bicarbonate = Moles of sodium chloride

4. Convert moles to mass of sodium bicarbonate:
Mass of sodium bicarbonate formed = Moles of sodium bicarbonate * Molar mass of sodium bicarbonate

Plug in the values and calculate to find the mass of sodium bicarbonate formed.

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