Answer:
The volume of carbon dioxide gas generated 468 mL.
Explanation:
The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%
Mass of tablet = 3.45 g
Mass of bicarbonate =[tex]3.45 g\times \frac{32.5}{100}=1.121 mol[/tex]
Moles of bicarbonate ion = [tex]\frac{1.121 g/mol}{61 g/mol}=0.01840 mol[/tex]
[tex]HCO_3^{-}(aq)+HCl(aq)\rightarrow H_2O(l)+CO_2(g)+Cl^-(aq)[/tex]
According to reaction, 1 mole of bicarbonate ion gives with 1 mole of carbon dioxide gas , then 0.01840 mole of bicarbonate ion will give:
[tex]\frac{1}{1}\times 0.01840 mol=0.01840 mol[/tex] of carbon dioxide gas
Moles of carbon dioxide gas n = 0.01840 mol
Pressure of the carbon dioxide gas = P = 1.00 atm
Temperature of the carbon dioxide gas = T = 37°C = 37+273 K=310 K
Volume of the carbon dioxide gas = V
[tex]PV=nRT[/tex] (ideal gas equation)
[tex]V=\frac{nRT}{P}=\frac{0.01840 mol\times 0.0821 atm L/mol K\times 310 K}{1.00 atm}=0.468 L[/tex]
1 L = 1000 mL
0.468 L =0.468 × 1000 mL = 468 mL
The volume of carbon dioxide gas generated 468 mL.
