Question

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calculate the volume of CO2 generated (in mL) at 37°C and 1.00 atm if a person were to accidentally ingest a 3.79-g tablet without following instructions. (Hint: The reaction occurs between HCO3− and HCl acid in the stomach.)

Answer 1

Answer : The volume of
`CO_2` will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

`HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2`

First we have to calculate the mass of
`HCO_3^-` in tablet.

`\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% * 3.79g=(32.5)/(100)* 3.79g=1.23175g`

Now we have to calculate the moles of
`HCO_3^-`.

Molar mass of
`HCO_3^-` = 1 + 12 + 3(16) = 61 g/mole

`\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=(1.23175g)/(61g/mole)=0.0202moles`

Now we have to calculate the moles of
`CO_2`.

From the balanced chemical reaction, we conclude that

As, 1 mole of
`HCO_3^-` react to give 1 mole of
`CO_2`

So, 0.0202 mole of
`HCO_3^-` react to give 0.0202 mole of
`CO_2`

The moles of
`CO_2` = 0.0202 mole

Now we have to calculate the volume of
`CO_2` by using ideal gas equation.

`PV=nRT`

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas =
`37^oC=273+37=310K`

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 :

`(1.00atm)* V=0.0202 mole* (0.0821L.atm/mole.K)* (310K)`

`V=0.51411L=514.11ml`

Therefore, the volume of
`CO_2` will be, 514.11 ml