Answer to Question #156164 in General Chemistry for Hargun

Question

Drinking a solution of baking soda, NaHCO3, can neutralize excess HCl(aq) in the stomach. A student stirred 5.0 g of baking soda in water and drank the solution. What is the size of “burp” expected from the neutralization reaction considering the products of the reaction are sodium chloride, water, and carbon dioxide gas. Assume the gas will be at a pressure of 1.03289 atm and a temperature of 37oC.

Answer 1

NaHCO₃ + HCl $\rightarrow$ NaCl + H₂O + CO₂(g)

No. of moles of baking soda taken = (5/84) moles which is equal to 0.0595 moles

According to stoichiometry whatever moles of NaHCO₃ reacts same no. of moles of CO₂ will be formed.

No. of moles of CO₂ formed = 0.0595 moles.

Using ideal gas equation:

PV = nRT

Temperature is given 310K (= 273+37 K)

Pressure = 1.03289 atm

R = 0.0821 Latm/K/mole

n = 0.0595 moles (moles of CO₂ produced)

So upon substituting these values in ideal gas equation and solve for value of Volume.

1.03289 × V = 0.0595 × 0.0821 × 310

we got V = 1.466 L