Question

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 3.45-g tablet without following instructions. (Hint: The reaction occurs between HCO3− and HCl acid in the stomach.)

Answer 1

The volume of carbon dioxide gas generated 468 mL.

Step-by-step explanation:

The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%

Mass of tablet = 3.45 g

Mass of bicarbonate =
`3.45 g* (32.5)/(100)=1.121 mol`

Moles of bicarbonate ion =
`(1.121 g/mol)/(61 g/mol)=0.01840 mol`

`HCO_3^(-)(aq)+HCl(aq)\rightarrow H_2O(l)+CO_2(g)+Cl^-(aq)`

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:

`(1)/(1)* 0.01840 mol=0.01840 mol` 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

`PV=nRT` (ideal gas equation)

`V=(nRT)/(P)=(0.01840 mol* 0.0821 atm L/mol K* 310 K)/(1.00 atm)=0.468 L`

1 L = 1000 mL

0.468 L =0.468 × 1000 mL = 468 mL

The volume of carbon dioxide gas generated 468 mL.