Question

a) The reaction volcano: Mix baking soda (sodium bicarbonate) with vinegar (acetic acid) to generate carbonic acid (further breaks into carbon dioxide and water) and sodium acetate. Write a balanced chemical equation for this reaction.b) I placed 4.15g of sodium bicarbonate (baking soda) into a 500mL flask. I then slowly added vinegar to the flask, waiting for the bubbles to subside before adding more acid. I continued to add vinegar until there were no more bubbles. Identify the limiting reactant in this reaction. Justify your choice.c) I placed the flask on a hotplate to boil off the remaining liquid until only the sodium acetate was left. Sodium acetate is a food preservative additive. The mass of the flask and product was 125.24 g. The empty flask weighed 122.60 g. Calculate the actual mass of the sodium acetate produced in this reaction.d) Based on the grams of sodium bicarbonate used (see step #2), how many grams of sodium acetate should be produced in this reaction? Show your work. e) Determine the percent yield of sodium acetate for this reaction. Show your work.

Answer 1

The question refers to the "volcano" reaction, which is a very common science experiment at school.

a) The "volcano reaction" consists in mixing baking soda (NaHCO3) with vinegar (CH3COOH) to produce carbon dioxide (CO2) and sodium acetate (CH3COONa).

The balanced reaction is:

`NaHCO_3\mleft(s\mright)+CH_3COOH\mleft(l\mright)\to CO_2\mleft(g\mright)+H_2O\mleft(l\mright)+CH_3COONa\mleft(aq\mright)`

b) In this case, we don't have the exact amount used for both reactants, but we can say that sodium bicarbonate (NaHCO3) is the limiting reactant because vinegar was added in excess. Another way to see it is: vinegar was added just to completely react with the available amount of baking soda, which makes the baking soda the limiting reactant.

c) The actual mass of sodium acetate obtained can be calculated from the difference between the masses of flask+sodium acetate (125.24g) and just flask (122.60g):

`m_(CH3COOH)=(125.24-122.60)g=2.64\text{g of sodium acetate}`

d) Considering the balanced reactin shown before and the amount of NaHCO3 used (4.15g), we can calculate the amount of sodium acetate that should be produced. But first, we need to know the amount of NaHCO3 used in moles (molar mass NaHCO3 = 84.00 g/mol):

`n_{NaHCO3_{}}=\frac{mass}{molar\text{ mass}}=\frac{4.15g}{84.00\text{ g/mol}}=0.0494\text{ mol of NaHCO}3`

Now, we can calculate the amount of CH3COONa that should be produced and convert it to mass of CH3COONa (molar mass = 82.03 g/mol):

1 mol NaHCO3 ---------- 1 mol CH3COOH

0.0494 mol NaHCO3 ----- x

Solving for x, we have that 0.0494 moles of CH3COONa or 4.052 g of CH3COONa should be produced.

e) Since we know the amount of CH3COONa that should be produced (4.052g) and the amount actually obtained (2.640g), we can calculate the yield of the reaction:

4.052g of CH3COONa ---------- 100% yield

2.640g of CH3COONa ---------- y

Solving for y, we have that the reaction had an yield of 65.2%.