Question

You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to fill this container given a 49.4% CO2 recovery, assuming the other contents in the bag take up a negligible volume compared to the gas. The temperature of the room is 24.5 °C and the atmospheric pressure is 753.5 mmHg.

Answer 1

3.636 grams of sodium bicarbonate is required.

Step-by-step explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

(
`1 atm = 760 mmHg`)

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65 K

Putting values in above equation, we get:

`(0.9914 atm)* 1.08 L=n* (0.0821L.atm/mol.K)* 297.65K\\\\n=0.0438 mole`

Percentage recovery of carbon dioxide gas = 49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

`(49.4)/(100)*  0.0438 mol=0.02164 mol`

`2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2`

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

`(2)/(1)* 0.02164 mol=0.04328 mol`

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.