Question

A \(41.41-\mathrm{mL}\) sample of a \(0.1250 \mathrm{M}\) acid reacts with an excess of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to form \(150.0 \mathrm{~mL} \mathrm{CO}_{2}\) at \(646 \mathrm{mmHg}\) and \(27^{\circ} \mathrm{C}\). If the acid is either \(\mathrm{HCl}\) or \(\mathrm{H}_{2} \mathrm{SO}_{4}\), which is it?

Short answer

The acid is \(\mathrm{H}_2\mathrm{SO}_4\) based on the 1:1 molar ratio with \(\mathrm{CO}_2\).

Step-by-step solution

Calculate Moles of Acid

First, we need to determine the moles of acid in the 41.41 mL sample. We convert mL to L by dividing by 1000, thus:\[ \text{Volume of acid in L} = \frac{41.41}{1000} = 0.04141 \text{ L} \]Then, using the molarity equation, \( M = \frac{\text{moles}}{\text{liters}} \), we calculate the moles:\[ \text{Moles of acid} = 0.1250 \, M \times 0.04141 \, L = 0.00517625 \, \text{moles} \]

Calculate Moles of CO2

Use the Ideal Gas Law to find the moles of \(\mathrm{CO}_{2}\). First, convert pressure to atmospheres by dividing by 760:\[ P = \frac{646}{760} = 0.850 \text{ atm} \]Convert temperature to Kelvin:\[ T = 27^{\circ}C + 273.15 = 300.15 \, K \]Convert volume from mL to L:\[ V = \frac{150.0}{1000} = 0.150 \, L \]Using the Ideal Gas Law \( PV = nRT \), solve for \(n\) (moles of \(\mathrm{CO}_{2}\)):\[ n = \frac{PV}{RT} = \frac{0.850 \, \text{atm} \times 0.150 \, \text{L}}{0.0821 \, \frac{\text{L atm}}{\text{K mol}} \times 300.15 \, \text{K}} = 0.00512 \, \text{moles} \]

Compare Moles of Acid and CO2

The number of moles of \(\mathrm{CO}_{2}\) is roughly equal to the number of moles of the acid from Step 1, suggesting a 1:1 stoichiometry. For \(\mathrm{HCl}\), the reaction is:\[ 2\, \mathrm{HCl} + \mathrm{Na}_2\mathrm{CO}_3 \rightarrow 2\, \mathrm{NaCl} + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \]For \(\mathrm{H}_2\mathrm{SO}_4\), the reaction is:\[ \mathrm{H}_2\mathrm{SO}_4 + \mathrm{Na}_2\mathrm{CO}_3 \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \]The \(\mathrm{H}_2\mathrm{SO}_4\) reaction has a 1:1 molar ratio with \(\mathrm{CO}_2\), confirming it aligns with our calculated moles.

Key concepts

Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It helps you figure out the right proportions of reactants and products in a chemical equation. For instance, baking cookies requires a specific amount of flour, sugar, and butter. Similarly, in chemistry, stoichiometry uses chemical equations to ensure that reactions are balanced and produce the desired amounts of products.

In our exercise, stoichiometry comes into play when we're comparing the moles of acid to the moles of carbon dioxide (\( \mathrm{CO}_2 \)) produced. With both \( \mathrm{HCl} \)and \( \mathrm{H}_2\mathrm{SO}_4 \), you need to look at the balanced chemical equations. Knowing the molar relationships allows you to deduce which acid was present based on the \( \mathrm{CO}_2 \)produced. Since the moles of \( \mathrm{CO}_2 \)matched closely with \( \mathrm{H}_2\mathrm{SO}_4 \), stoichiometry helped identify the acid.

Acid-Base Reactions

Acid-base reactions are a type of chemical reaction where an acid reacts with a base to produce salt and usually water. This is a fundamental concept in chemistry when studying reactions between different compounds.

In the given problem, the acid (\( \mathrm{HCl} \)or \( \mathrm{H}_2\mathrm{SO}_4 \)) reacts with the base, sodium carbonate (\( \mathrm{Na}_2\mathrm{CO}_3 \)). Despite being a bit more complex than simple acid-base neutralization (as it produces gas), the idea is the same. The acid donates protons, while the carbonate serves as the base, accepting protons and releasing \( \mathrm{CO}_2 \) gas as a byproduct. Understanding this helps explain the transformation of reactants to products and why certain products like \( \mathrm{CO}_2 \) were formed.

Molarity

Molarity is an important concept in chemistry, representing the concentration of a solution. It is defined as the number of moles of solute per liter of solution and is often denoted by the symbol \( M \). Molarity is crucial when performing calculations involving solutions and reactions.

In our problem, molarity is used to find the amount of moles of acid in the given sample. Using the formula \( M = \frac{\text{moles}}{\text{liters}} \), we calculated the moles of acid present in 41.41 mL of a \( 0.1250 \mathrm{M} \) solution. This step is essential as it sets up further calculations, like using the Ideal Gas Law, to determine the identity of the acid based on the reaction products.

Chemical Reactions

Chemical reactions are processes in which substances, known as reactants, are transformed into different substances, known as products. Understanding chemical reactions involves recognizing how substances interact, rearrange, and change bonds.

In the exercise, two potential reactions were analyzed. Both involve an acid and sodium carbonate, leading to the production of carbon dioxide gas (\( \mathrm{CO}_2 \)). The reactions showed different stoichiometry, depending on whether the reactant was \( \mathrm{HCl} \)or \( \mathrm{H}_2\mathrm{SO}_4 \). Identifying the correct chemical reaction helped determine the acid used. Remember, the balance in chemical equations is key. This maintains the conservation of mass, ensuring no atoms are lost or gained, just rearranged into new products.