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A \(21.4-\mathrm{mL}\) volume of hydrochloric acid reacts completely with a solid sample of \(\mathrm{MgCO}_{3} .\) The reaction is \(2 \mathrm{HCl}(a q)+\mathrm{MgCO}_{3}(s) \longrightarrow\) $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O (l)+\mathrm{MgCl}_{2}(a q) $$ The volume of \(\mathrm{CO}_{2}\) formed is \(159 \mathrm{~mL}\) at \(23^{\circ} \mathrm{C}\) and \(731 \mathrm{mmHg} .\) What is the molarity of the \(\mathrm{HCl}\) solution?

Short Answer

Expert verified
The molarity of the \(\mathrm{HCl}\) solution is approximately \(0.600\, \mathrm{M}\).

Step by step solution

01

Write the Balanced Equation

The given reaction is already balanced:\[2 \mathrm{HCl}(aq) + \mathrm{MgCO}_{3}(s) \rightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2}O(l) + \mathrm{MgCl}_{2}(aq)\].This shows that 2 moles of \(\mathrm{HCl}\) react with 1 mole of \(\mathrm{MgCO}_{3}\) to produce 1 mole of \(\mathrm{CO}_{2}\).
02

Convert CO2 Volume to Moles

Use the ideal gas law to find moles of \(\mathrm{CO}_{2}\):\[PV = nRT\]Convert \(23^{\circ} \mathrm{C}\) to Kelvin: \[T = 23 + 273.15 = 296.15\, \mathrm{K}\]Convert pressure from mmHg to atm: \[P = \frac{731}{760} \approx 0.9618\, \mathrm{atm}\]The volume \(V\) is \(0.159\, \mathrm{L}\). Using \(R = 0.0821\, \mathrm{L\, atm/(mol\, K)}\), solve for \(n\):\[n = \frac{PV}{RT} = \frac{0.9618 \times 0.159}{0.0821 \times 296.15} \approx 0.00642\, \mathrm{mol}\, \mathrm{CO}_{2}\].
03

Relate Moles of HCl to CO2

From the balanced equation, \(2\) moles of \(\mathrm{HCl}\) produce \(1\) mole of \(\mathrm{CO}_{2}\). Therefore, the moles of \(\mathrm{HCl}\) that reacted are:\[2 \times 0.00642 = 0.01284\, \mathrm{mol}\, \mathrm{HCl}\]
04

Calculate Molarity of HCl Solution

Molarity \(M\) is calculated by dividing the moles of solute by the volume of solution in liters:\[M = \frac{\text{moles of } \mathrm{HCl}}{\text{volume of solution in L}} = \frac{0.01284}{0.0214} \approx 0.600\, \mathrm{M}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity
Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. Here’s a formula to represent molarity:\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]In the exercise above, you needed to find the molarity of hydrochloric acid (HCl) based on its reaction with magnesium carbonate (\(\text{MgCO}_3\)).
The reaction resulted in carbon dioxide (\(\text{CO}_2\)) being formed, and the volume of the hydrochloric acid solution was known. By calculating the moles of HCl that participated in the reaction, you could then find the molarity using the formula above.
Molarity helps chemists understand how much reactant is present in a given volume of solution, which is critical for predicting how reactions will proceed and for ensuring that the reactions occur under controlled conditions.
Chemical Reactions
Chemical reactions involve transforming one or more substances into different substances. During a reaction, chemical bonds break and new ones form. In the provided exercise, the chemical reaction involves hydrochloric acid reacting with magnesium carbonate.
  • Reactants: These are substances that start off a reaction. In this case, the reactants are hydrochloric acid (\(\text{HCl}\)) and magnesium carbonate (\(\text{MgCO}_3\)).
  • Products: These are the substances formed as a result of the reaction. Here, the products are carbon dioxide (\(\text{CO}_2\)), water (\(\text{H}_2\text{O}\)), and magnesium chloride (\(\text{MgCl}_2\)).
  • Reaction conditions: These often include temperature and pressure, as seen in the problem where the temperature of \(23^\circ\text{C}\) and pressure of \(731\text{ mmHg}\) are given for the formation of \(\text{CO}_2\).
Understanding chemical reactions and their conditions is key in the study of chemistry, allowing scientists to predict products, rates, and energy changes.
Stoichiometry
Stoichiometry is a branch of chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the conservation of mass and the concept of moles. Stoichiometry enables the calculation of quantities needed or produced in a reaction.In the chemical reaction between hydrochloric acid and magnesium carbonate, stoichiometry is used to determine how much of each reactant is needed to produce a certain amount of product. The balanced chemical equation:\[2 \text{HCl}(aq) + \text{MgCO}_3(s) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l) + \text{MgCl}_2(aq)\]Indicates that two moles of \(\text{HCl}\) react with one mole of \(\text{MgCO}_3\). It also shows how calculations are made to understand how reactants transform into products by using ratios derived from the balanced equation.
Stoichiometry is essential for predicting the exact amounts of substances consumed and produced, allowing scientists and engineers to scale reactions to desired proportions safely and effectively.
Balanced Chemical Equation
A balanced chemical equation represents a chemical reaction with the same number of each type of atom on both sides of the equation. Balancing ensures that mass is conserved throughout the reaction, as dictated by the law of conservation of mass. In the exercise, the given balanced equation is:\[2 \text{HCl}(aq) + \text{MgCO}_3(s) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l) + \text{MgCl}_2(aq)\]The equation shows that two molecules of hydrochloric acid react with one molecule of magnesium carbonate to produce one molecule of carbon dioxide, one molecule of water, and one molecule of magnesium chloride.
This balance is crucial because it allows chemists to predict how much of a chemical is required and how much product will form, ensuring that quantities involved in a reaction are correctly accounted. It helps in maximizing efficiency and safety in chemical processes. Understanding and using balanced equations is a fundamental skill in chemistry.

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Most popular questions from this chapter

A McLeod gauge measures low gas pressures by compressing a known volume of the gas at constant temperature. If \(315 \mathrm{~cm}^{3}\) of gas is compressed to a volume of 0.0457 \(\mathrm{cm}^{3}\) under a pressure of \(2.51 \mathrm{kPa}\), what was the original gas pressure?

5.161 Power plants driven by fossil-fuel combustion generate substantial greenhouse gases (e.g., \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ) as well as gases that contribute to poor air quality (e.g., \(\mathrm{SO}_{2}\) ). To evaluate exhaust emissions for regulation purposes, the generally inert nitrogen supplied with air must be included in the balanced reactions; it is further assumed that air is composed of only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) with a volume/volume ratio of \(3.76: 1.00,\) respectively. In addition to the water produced, the fuel's \(\mathrm{C}\) and \(\mathrm{S}\) are converted to \(\mathrm{CO}_{2}\) and \(\mathrm{SO}_{2}\) at the stack. Given these stipulations, answer the following for a power plant running on a fossil fuel having the formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\). Including the \(\mathrm{N}_{2}\) supplied in air, write a balanced combustion equation for the complex fuel assuming \(120 \%\) stoichiometric combustion (i.e., when excess oxygen is present in the product gases). Except for \(\mathrm{N}_{2},\) use only integer coefficients. (See problem \(3.141 .\) ) What gas law is used to effectively convert the given \(\mathrm{N}_{2} / \mathrm{O}_{2}\) volume ratio to a molar ratio when deriving the balanced combustion equation in (a)? (i) Boyle's law (ii) Charles's law (iii) Avogadro's law (iv) ideal gas law c) Assuming the product water condenses, use the result from (b) to determine the stack gas composition on a "dry" basis by calculating the volume/volume percentages for \(\mathrm{CO}_{2}, \mathrm{~N}_{2},\) and \(\mathrm{SO}_{2}\) Assuming the product water remains as vapor, repeat (c) on a "wet" basis.Assuming the product water remains as vapor, repeat (c) on a "wet" basis. Some fuels are cheaper than others, particularly those with higher sulfur contents that lead to poorer air quality. In port, when faced with ever-increasing air quality regulations, large ships operate power plants on more costly, higher-grade fuels; with regulations nonexistent at sea, ships' officers make the cost-saving switch to lower-grade fuels. Suppose two fuels are available, one having the general formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and the other \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S} .\) Based on air quality concerns alone, which fuel is more likely to be used when idling in port? Support your answer to (d) by calculating the mass ratio of \(\mathrm{SO}_{2}\) produced to fuel burned for \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S},\) assuming \(120 \%\) stoichiometric combustion.

A flask contains \(201 \mathrm{~mL}\) of argon at \(21^{\circ} \mathrm{C}\) and \(738 \mathrm{mmHg}\). What is the volume of gas, corrected to STP?

A \(275-\mathrm{mL}\) sample of CO gas is collected over water at \(31^{\circ} \mathrm{C}\) and \(755 \mathrm{mmHg}\). If the temperature of the gas collection apparatus rises to \(39^{\circ} \mathrm{C}\), what is the new volume of the sample? Assume that the barometric pressure does not change.

Gas Laws and Kinetic Theory of Gases I Shown here are two identical containers labeled \(\mathrm{A}\) and \(\mathrm{B}\). Container A contains a molecule of an ideal gas, and container B contains two molecules of an ideal gas. Both containers are at the same temperature. (Note that small numbers of molecules and atoms are being represented in these examples in order that you can easily compare the amounts. Real containers with so few molecules and atoms would be unlikely.) How do the pressures in the two containers compare? Be sure to explain your answer. Shown below are four different containers \((\mathrm{C}, \mathrm{D}, \mathrm{E}\) and \(\mathrm{F}\) ), each with the same volume and at the same temperature. How do the pressures of the gases in the containers compare? Container \(\mathrm{H}\) below has twice the volume of container G. How will the pressure in the containers compare? Explain your reasoning. How will the pressure of containers \(\mathrm{G}\) and \(\mathrm{H}\) compare if you add two more gas molecules to container \(\mathrm{H}\) ? Consider containers I and J below. Container J has twice the volume of container \(\mathrm{I}\). Container \(\mathrm{I}\) is at a temperature of \(100 \mathrm{~K},\) and container \(\mathrm{J}\) is at \(200 \mathrm{~K}\). How does the pressure in container I compare with that in container \(\mathrm{J} ?\) Include an explanation as part of your answer.

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