Question

The percent by mass of bicarbonate $\left({\mathrm{HCO}}_3^{-}\right)$ in a certain Alka-Seltzer product is 32.5 percent. Calculate the volume of ${\mathrm{CO}}_2$ generated (in $\mathrm{mL}$ ) at ${37}^{\circ }\mathrm{C}$ and 1.00 atm when a person ingests a 3.29 -g tablet. (Hint: The reaction is between ${\mathrm{HCO}}_3^{-}$ and $\mathrm{HCl}$ acid in the stomach.)

Short answer

The volume of CO2 generated is approximately 446 mL.

Step-by-step solution

Determine the Mass of Bicarbonate

Given that bicarbonate makes up 32.5% of the Alka-Seltzer tablet, calculate the mass of bicarbonate in a 3.29 g tablet:$${\text{Mass of HCO}}_3^{-}=0.325\times 3.29\text{g}=1.06925\text{g}$$

Convert Bicarbonate Mass to Moles

Find the molar mass of bicarbonate, ${\text{HCO}}_3^{-}$, which is approximately 61.01 g/mol. Convert the mass of bicarbonate to moles:$${\text{Moles of HCO}}_3^{-}=\frac{1.06925\text{g}}{61.01\text{g/mol}}\approx 0.01753\text{mol}$$

Determine the Moles of CO2 Produced

The chemical reaction between bicarbonate ${\text{HCO}}_3^{-}$ and hydrochloric acid $\text{HCl}$ in the stomach produces carbon dioxide ${\text{CO}}_2$. The balanced chemical equation is: $${\text{HCO}}_3^{-}+\text{HCl}\to {\text{CO}}_2+{\text{H}}_2\text{O}+{\text{Cl}}^{-}$$This shows a 1:1 mole ratio between ${\text{HCO}}_3^{-}$ and ${\text{CO}}_2$. Therefore, the moles of ${\text{CO}}_2$ produced are the same as the moles of ${\text{HCO}}_3^{-}$: $${\text{Moles of CO}}_2=0.01753\text{mol}$$

Use Ideal Gas Law to Calculate Volume of CO2

Apply the ideal gas law, $PV=nRT$, where:- $P=1.00\text{atm}$- $n=0.01753\text{mol}$- $R=0.0821{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}$- $T=273.15+37.0=310.15\text{ K}$Solve for $V$:$$V=\frac{nRT}{P}=\frac{0.01753\times 0.0821\times 310.15}{1.00}\approx 0.446\text{ L}$$Since the final answer is required in milliliters, convert liters to milliliters:$$V=0.446\text{ L}\times 1000\text{ mL/L}=446\text{ mL}$$

Key concepts

Percent by Mass

Percent by mass is an important concept in chemistry. It helps to express the amount of a particular element or compound in a mixture or solution. Imagine it as the 'piece of the pie' when you know the whole pie's size! To calculate it, you use a simple formula:

• First, take the mass of the element or compound we're interested in.
• Then, divide it by the total mass of the mixture or solution.
• Finally, multiply the result by 100 to convert it to a percentage.

For example, in the exercise, bicarbonate ($ext{HCO}_3^{-}$) makes up 32.5% of an Alka-Seltzer tablet. From here, the mass of bicarbonate in a 3.29-gram tablet is easy to find. Simply multiply 3.29 g by 32.5% (or 0.325 in decimal form). This gives you 1.06925 g of bicarbonate in the mixture. This kind of calculation is crucial in understanding the composition of compounds in mixtures, especially in medicine or dietary adjustments.

Moles and Molar Mass

The concept of moles and molar mass is fundamental in chemistry, making it possible to connect the macroscopic world of materials we see to the microscopic world of atoms and molecules. The mole, a unit of measurement, links the mass of a substance to its quantity in particles. A mole of any element or compound contains Avogadro's number of particles, approximately 6.022 x 10²³ particles.
To understand how many moles you have, you use the molar mass, which is the mass of one mole of a substance. This information is often found on the periodic table for elements and can be calculated for compounds. For bicarbonate (HCO₃^-), the molar mass is approximately 61.01 g/mol.
The exercise uses this principle by dividing the mass of bicarbonate from the tablet (1.06925 g) by its molar mass (61.01 g/mol), resulting in about 0.01753 moles of bicarbonate. This conversion is essential because chemical reactions don’t happen with grams, they happen with molecules. So, moles allow us to count those molecules!

Chemical Reactions

Chemical reactions are the processes by which substances transform into new substances. They involve breaking and forming chemical bonds, resulting in one or more substances changing identity. In the context of the exercise, the reaction involves bicarbonate (HCO₃^-) reacting with the hydrochloric acid (HCl) in your stomach.
The balanced equation for this reaction is:$${\text{HCO}}_3^{-}+\text{HCl}\to {\text{CO}}_2+{\text{H}}_2\text{O}+{\text{Cl}}^{-}$$This equation tells us a lot:

• The reactants, HCO₃^- and HCl, combine in a 1:1 ratio, meaning one mole of bicarbonate reacts with one mole of hydrochloric acid.
• The products formed are carbon dioxide (CO₂), water (H₂O), and chloride ions (Cl^-).

Understanding the stoichiometry, or the quantitative aspects of these reactions, allows us to predict how much of each product is formed. In the exercise, since the moles of bicarbonate were found to be 0.01753 mol, it's a straightforward prediction that 0.01753 mol of CO₂ will be produced. Chemical reactions help us grasp how materials interact and change, forming the basis for everything from digestion to chemical manufacturing.

Volume Conversions

Volume conversions allow us to switch between various units of measurement, which is especially important in scientific calculations. In chemistry, often we deal with gases, and their volumes are typically measured in liters or milliliters. To convert, you use the fact that 1 liter equals 1000 milliliters.
In the exercise, once the volume of produced CO₂ was calculated using the Ideal Gas Law (resulting in 0.446 L), it required a simple conversion from liters to milliliters to fit the problem's requirement.

• You multiply 0.446 L by 1000 ml/L to convert it to milliliters, getting 446 ml.

Understanding conversions like this is crucial, not just for calculations, but for real-world applications in medicine, cooking, and engineering. By managing different measurement units seamlessly, we can ensure accuracy and efficiency in science and daily life.