Answer:
3.636 grams of sodium bicarbonate is required.
Explanation:
Using ideal gas equation:
PV = nRT
where,
P = Pressure of gas = 753.5 mmHg = 0.9914 atm
(
)
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)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole](https://tex.z-dn.net/?f=%280.9914%20atm%29%5Ctimes%201.08%20L%3Dn%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20297.65K%5C%5C%5C%5Cn%3D0.0438%20mole)
Percentage recovery of carbon dioxide gas = 49.4%
Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole
![\frac{49.4}{100}\times 0.0438 mol=0.02164 mol](https://tex.z-dn.net/?f=%5Cfrac%7B49.4%7D%7B100%7D%5Ctimes%20%200.0438%20mol%3D0.02164%20mol)
![2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2](https://tex.z-dn.net/?f=2NaHCO_3%5Crigharrow%20Na_2CO_3%2BH_2O%2BCO_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:
![\frac{2}{1}\times 0.02164 mol=0.04328 mol](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B1%7D%5Ctimes%200.02164%20mol%3D0.04328%20mol)
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.