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

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 =$3.45 g\times \frac{32.5}{100}=1.121 mol$

Moles of bicarbonate ion = $\frac{1.121 g/mol}{61 g/mol}=0.01840 mol$

$HCO_3^{-}(aq)+HCl(aq)\rightarrow H_2O(l)+CO_2(g)+Cl^-(aq)$

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:

$\frac{1}{1}\times 0.01840 mol=0.01840 mol$ 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)

$V=\frac{nRT}{P}=\frac{0.01840 mol\times 0.0821 atm L/mol K\times 310 K}{1.00 atm}=0.468 L$

1 L = 1000 mL

0.468 L =0.468 × 1000 mL = 468 mL

The volume of carbon dioxide gas generated 468 mL.