You determine the volume of your plastic bag (simulated human stomach) is 1.05 l.

how many grams of nahco3 (s) are required to fill this container given a 70.2% co2 recovery, assuming the other contents in the bag take up a negligible volume compared to the gas.

the temperature of the room is 21.5 â°c and the atmospheric pressure is 753.5 mmhg.

We need 5.1 grams of NaHCO3

Explanation:

Step 1: Data given

volume of the plactic bag = 1.05 L

Recovery = 70.2 %

Temperature = 21.5°C

Pressure = 753.5 mmHg

Step 2: ideal gas law

p*V = n*R*T

with p = the pressure = 735.5 mmHg = 0.9914 atm

with V= the volume = 1.05 L

with n = the number of moles =  unknown

with R =  Gas constant = 0.0821 L*atm/mol*K

with T =the temperature = 21.5 °C = 296.65 Kelvin

Calculate the number of moles:

n = p*V / RT

n = (0.9914 * 1.05)/(0.0821 * 296.65)

n =0.0427 mol

Percentage recovery of carbon dioxide gas =  70.2%

Step 3: Calculate actual moles of CO2 formed

Actual moles of carbon dioxide formed: 70.2 % of 0.0427 mol

0.702 * 0.0427 = 0.0300 mol

Step 4: Calculate moles NaHCO3

2NaHCO3 → Na2CO3 + H2O + CO2

For 2 mole NaHCO3 consumed, we get 1 mole Na2CO3, 1 mole H2O and 1 mole CO2

2*0.0300 = 0.06 mol

Step 5: Calculate mass of NaHCO3

Calculate mass of NaHCO3

Mass NaHCO3 = moles NaHCO3 * Molar mass NaHCO3

Mass NaHCO3 = 0.06 mol * 84 g/mol

Mass NaHCO3 = 5.1 grams

We need 5.1 grams of NaHCO3