edit*** Be sure to place your 600 mL beaker now in its proper location in the la

Question

edit*** Be sure to place your 600 mL beaker now in its proper location in the lab setup. You can anchor it with a
piece of tape if needed. Just be careful the beaker doesn’t tip over or the tube doesn’t fall out.
9. Determine the barometric pressure (in atm) for your specific location (look it up online if you don’t have a
barometer lying around), and record it in Table 1 (see below).
10. Place a piece of weigh paper on the scale, and tare (re-zero) the scale.
11. Weight out approximately 1.0 g of sodium bicarbonate (NaHCO3) onto the weigh paper. It doesn’t have to be
exactly 1.0 g, just get it as close as you can, and be sure to record the actual value (in grams) in Table 1 (see
below).
12. Carefully remove the stopper from Flask 1 and add 25.0 mL of 4.5% acetic acid.
13. Quickly transfer your sodium bicarbonate into the flask with the acetic acid in Flask 1 and replace the stopper
immediately.
14. Allow the reaction to react to completion, being sure to collect all the displaced water in the 600 mL beaker.
Remember, as was explained in the last lab, the mL of water collected is equivalent to the mL of gas
produced in the experiment.

Explanation / Answer

CHCOOH (aq) + NaHCO3 (s) --------> CHCOONa (aq) + CO(g) + HO(l)

The moles of CO2 is found from the gas law, V = n*R*T/P

For trial 1.

V = volume of CO2 in L = water displaced in L = 0.072L

T = temp in K = 298.45K

P = pressure in atm = 1 atm

R = 0.08207 L-atm/ºK-mol

n = V * P /RT

n = 0.072L * 1atm /0.08207 L-atm/ºK-mol *298.45K

n = 0.072/24.493moles

n= 0.00293 moles CO2

1 moles of CO2 is produced from 1moles of NaHCO

0.00293 moles of CO2 is produced from 00293moles of NaHCO

Molecular weight of NaHCO = 84g/mole

So g of NaHCO in 0.00293moles = 0.00293 moles *84 g /mole = 0.246g

For trial 2.similarly we calculate the moles of CO2 produced

n = V * P /RT

n = 0.073L * 1atm /0.08207 L-atm/ºK-mol *298.25K

n = 0.073/24.477moles

n= 0.00298 moles CO2

0.00298 moles of CO2 is produced from0. 00298moles of NaHCO

So g of NaHCO in 0.00298moles = 0.00298 moles *84 g /mole = 0.250g

For trial 3.

Na2 CO3 + CH3COOH CH3COONa + H2O + CO2 (g)

n = V * P /RT

n = 0.098L * 1atm /0.08207 L-atm/ºK-mol *298.45K

n = 0.098/24.493moles

n= 0.00293 moles CO2

1 moles of CO2 is produced from 1moles of Na2CO

0.00293 moles of CO2 is produced from 00293moles of Na2 CO

Molecular weight of Na2 CO = 60g/mole

So g of Na2CO in 0.00293moles = 0.00293 moles *60 g /mole = 0.176g

For trial 4.

n = V * P /RT n = 0.105L * 1atm /0.08207 L-atm/ºK-mol *298.45K

n = 0.105/24.493moles

n= 0.00428 moles CO2

1 moles of CO2 is produced from 1moles of Na2CO

0.00429 moles of CO2 is produced from 00428 moles of Na2 CO

So g of Na2CO in 0.00429 moles = 0.00429 moles *60 g /mole = 0.257g

Trial

Pressure atm

Temperature K

Reactant mass in g

Water displaced L

1(NaHCO )

1

298.45

0.246

0.072

2 (NaHCO )

1

298.25

0.250

0.073

3 (Na2CO)

1

298.45

0.176

0.098

4 (Na2CO)

1

298.45

0.257

0.105

Trial

Pressure atm

Temperature K

Reactant mass in g

Water displaced L

1(NaHCO )

1

298.45

0.246

0.072

2 (NaHCO )

1

298.25

0.250

0.073

3 (Na2CO)

1

298.45

0.176

0.098

4 (Na2CO)

1

298.45

0.257

0.105

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.