Respuesta :
Answer : The volume of [tex]CO_2[/tex] will be, 514.11 ml
Explanation :
The balanced chemical reaction will be,
[tex]HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2[/tex]
First we have to calculate the mass of [tex]HCO_3^-[/tex] in tablet.
[tex]\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g[/tex]
Now we have to calculate the moles of [tex]HCO_3^-[/tex].
Molar mass of [tex]HCO_3^-[/tex] = 1 + 12 + 3(16) = 61 g/mole
[tex]\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex].
From the balanced chemical reaction, we conclude that
As, 1 mole of [tex]HCO_3^-[/tex] react to give 1 mole of [tex]CO_2[/tex]
So, 0.0202 mole of [tex]HCO_3^-[/tex] react to give 0.0202 mole of [tex]CO_2[/tex]
The moles of [tex]CO_2[/tex] = 0.0202 mole
Now we have to calculate the volume of [tex]CO_2[/tex] by using ideal gas equation.
[tex]PV=nRT[/tex]
where,
P = pressure of gas = 1.00 atm
V = volume of gas = ?
T = temperature of gas = [tex]37^oC=273+37=310K[/tex]
n = number of moles of gas = 0.0202 mole
R = gas constant = 0.0821 L.atm/mole.K
Now put all the given values in the ideal gas equation, we get :
[tex](1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)[/tex]
[tex]V=0.51411L=514.11ml[/tex]
Therefore, the volume of [tex]CO_2[/tex] will be, 514.11 ml
