information about some commonly sold plastic Zi Below is a table the volume of g

ID: 548512 • Letter: I

Question

information about some commonly sold plastic Zi Below is a table the volume of gas required to inflate each bag fully, and ts o to bags. Calculate then use the ideal gas law to calculate the and 6.0 M acetic acid, respectively, needed to ornduce that volume of gas. Sum bicarbonates fully, l your work in the space below the table and on the back of produce that volume of gas. Sh °C, Pressure - 740 mm H this sheet. Temperature 20 c Moles CO2 Bag Volume I Pint I Quart Gallon Grams of NaHCO3 mL CH3OOH

Explanation / Answer

Assume that CO2 fills up the Ziploc bags completely so that the volume of the gas is equal to the volume of the bag.

The following conversion factors are essential:

1 pint = 473.176 mL

1 quart = 946.353 mL

1 gallon = 3785.41 mL

Pressure of the gas, P = 740 mmHg = (740 mmHg)*(1 atm/760 mmHg) = 0.9737 atm.

Temperature of the gas, T = 20°C = (20 + 273) K = 293 K.

Molar mass of NaHCO3 = 84.007 g/mol

Molar mass of CH3COOH = 60.05 g/mol

We will use the ideal gas law: P*V = n*R*T where V is the volume of the gas in liters; n = number of moles of the gas. Therefore,

n = P*V/RT

Fill in the first part of the table as below.

Volume of gas

Volume of gas in L

Number of moles of CO2, n = P*V/RT

1 pint

(473.176 mL)*(1 L/1000 mL) = 0.473176 L

n = (0.9737 atm)*(0.473176 L)/(0.082 L-atm/mol.K).(293 K) = 0.01918 mole 0.019 mole

1 quart

(946.353 mL)*(1 L/1000 mL) = 0.946353 L

n = (0.9737 atm)*(0.946353 L)/(0.082 L-atm/mol.K).(293 K) = 0.03835 mole 0.038 mole

1 gallon

(3785.41 mL)*(1 L/1000 mL) = 3.78541 L

n = (0.9737 atm)*(3.78541 L)/(0.082 L-atm/mol.K).(293 K) = 0.15341 mole 0.153 mole

Next fill up the second part of the table as below.

Moles CO2

Moles NaHCO3

Grams NaHCO3

Moles CH3COOH

Grams CH3COOH

0.019

0.038 (check sample calculation 1 below)

3.1923 (check sample calculation 2 below)

0.019 (check sample calculation 3 below)

1.1409 (check sample calculation 4 below)

0.038

0.076

6.3845

0.038

2.2819

0.153

0.306

25.7061

0.153

9.1876

Sample Calculation 1:

Write down the decomposition of NaHCO3 as below.

2 NaHCO3 -------> Na2CO3 + CO2 + H2O

As per the stoichiometric equation,

2 mole NaHCO3 = 1 mole CO2.

Therefore,

0.019 mole CO2 = (0.019 mole CO2)*(2 mole NaHCO3/1 mole CO2) = 0.038 mole NaHCO3.

Sample Calculation 2:

Mass of NaHCO3 required = (0.038 mole)*(84.007 g/mol) = 3.192266 g 3.1923 g.

Sample Calculation 3:

CO2 is obtained by the oxidation of acetic acid as below.

CH3COOH -------> CO2 + 2 H2O

As per the stoichiometric equation,

1 mole CO2 = 1 mole CH3COOH.

Therefore,

0.019 mole CO2 = (0.019 mole CO2)*(1 mole CH3COOH/1 mole CO2) = 0.019 mole CH3COOH.

Sample Calculation 4:

Mass of CH3COOH required = (0.019 mole)*(60.05 g/mol) = 1.14095 g 1.1409 g.