Respuesta :
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
([tex] 1 atm = 760 mmHg[/tex])
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:
[tex](0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole[/tex]
Percentage recovery of carbon dioxide gas = 49.4%
Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole
[tex]\frac{49.4}{100}\times 0.0438 mol=0.02164 mol[/tex]
[tex]2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2[/tex]
According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.
Then 0.02164 moles f carbon dioxide will be obtained from:
[tex]\frac{2}{1}\times 0.02164 mol=0.04328 mol[/tex]
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.
