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Magnesium metal reacts with hydrochloric acid to produce hydrogen gas, \(\mathrm{H}_{2}\). $$ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ Calculate the volume (in liters) of hydrogen produced at \(33^{\circ} \mathrm{C}\) and \(665 \mathrm{mmHg}\) from \(0.0840 \mathrm{~mol} \mathrm{Mg}\) and excess \(\mathrm{HCl}\).

Short Answer

Expert verified
The volume of hydrogen gas produced is approximately 2.43 liters.

Step by step solution

01

Write the Balanced Chemical Equation

The balanced chemical equation for the reaction is: \[ \mathrm{Mg}(s) + 2 \mathrm{HCl}(aq) \rightarrow \mathrm{MgCl}_{2}(aq) + \mathrm{H}_{2}(g) \] This equation already indicates that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.
02

Determine Moles of Hydrogen Gas

From the balanced equation, it's clear that 1 mole of \( \mathrm{Mg} \) produces 1 mole of \( \mathrm{H}_{2} \). Thus, \( 0.0840 \) moles of \( \mathrm{Mg} \) will produce \( 0.0840 \) moles of \( \mathrm{H}_{2} \) gas.
03

Convert Temperature to Kelvin

The temperature needs to be in Kelvin for gas law calculations. Convert \(33^{\circ} \mathrm{C}\) to Kelvin by adding 273.15. Therefore, \[ T = 33 + 273.15 = 306.15 \text{ K} \].
04

Convert Pressure to Atmospheres

The pressure given is in mmHg and needs to be converted to atmospheres. Use the conversion: \( 1 \text{ atm} = 760 \text{ mmHg} \). Thus, \[ P = \frac{665 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.875 \text{ atm} \].
05

Use the Ideal Gas Law to Find Volume

Use the ideal gas law formula \( PV = nRT \), where \( P \) is the pressure in atm, \( V \) is the volume in liters, \( n \) is the moles of gas, \( R \) is the gas constant \( 0.0821 \text{ L atm/mol K} \), and \( T \) is the temperature in Kelvin. Substitute the known values: \[ 0.875 \cdot V = 0.0840 \times 0.0821 \times 306.15 \].
06

Calculate the Volume

Rearrange the ideal gas law formula to solve for \( V \):\[ V = \frac{0.0840 \times 0.0821 \times 306.15}{0.875} \approx 2.43 \text{ L} \]. Therefore, the volume of hydrogen gas produced is approximately 2.43 liters.

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Key Concepts

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

Chemical Reaction
A chemical reaction occurs when substances, known as reactants, transform into different substances, called products. In this exercise, magnesium metal (Mg) reacts with hydrochloric acid (HCl). The products of this reaction are magnesium chloride (MgCl extsubscript{2}) and hydrogen gas (H extsubscript{2}).
This is an example of a single-replacement reaction. The magnesium replaces the hydrogen ions in the hydrochloric acid to form the magnesium chloride salt and hydrogen gas.
Understanding chemical reactions is crucial as they are fundamental processes found in nature and industrial applications. Without such reactions, the transformation of materials and energy changes necessary for life would not be possible.
  • Reactants: The starting substances, Mg and HCl.
  • Products: The new substances created, MgCl extsubscript{2} and H extsubscript{2}.
  • Balanced Equation: It's crucial to ensure that the number of atoms for each element is the same on both sides of the equation, maintaining mass conservation.
Stoichiometry
Stoichiometry is a concept in chemistry that involves the calculation of reactants and products in chemical reactions. It uses the coefficients of a balanced chemical equation to determine the ratios of the substances involved.
In the given exercise, stoichiometry helps us understand the relationship between magnesium and hydrogen gas. The balanced equation shows us that one mole of magnesium reacts with two moles of hydrochloric acid to produce one mole of hydrogen gas.
This stoichiometric relationship allows us to determine the amount of any other substance in the reaction once one quantity is known.
  • The ratio from the balanced equation: 1 Mg : 2 HCl : 1 H extsubscript{2}.
  • If you have 0.0840 moles of Mg, it will produce 0.0840 moles of H extsubscript{2}, assuming HCl is in excess.
Stoichiometry is a powerful tool in predicting yields and proportions in chemistry and is essential for any quantitative analysis of reactions.
Gas Laws
The gas laws are a set of laws that describe how gases behave under various conditions. They are essential for predicting how gases will respond to changes in temperature, pressure, and volume.
The most relevant gas law here is the Ideal Gas Law, stated as \( PV = nRT \). This formula allows us to calculate the volume of a gas when its pressure, number of moles, and temperature are known.
For the exercise, we must convert the given conditions to the appropriate units to use the Ideal Gas Law effectively.
  • Pressure needs to be in atmospheres (atm).
  • Temperature must be in Kelvin (K).
  • The gas constant \( R \) is 0.0821 L atm/mol K.
This exercise shows how the Ideal Gas Law helps calculate the volume of hydrogen gas produced from the reaction at specific conditions.
Moles of Gas
Moles are a fundamental unit in chemistry that measure the quantity of any given substance. In gaseous reactions, particularly, they are crucial for calculating other parameters like volume or pressure.
The Ideal Gas Law, \( PV = nRT \), demonstrates how the moles of gas \( n \) directly relate to changes in volume, pressure, or temperature.
In the exercise, knowing the moles of magnesium (0.0840 moles) allows us to determine the moles of hydrogen gas produced, which are also 0.0840 moles, as per the balanced chemical equation.
This helps in using the Ideal Gas Law to calculate the volume of gas produced. Understanding moles and their conversions is vital in chemistry because they act as a bridge between the atomic scale and the real-world scale in chemical calculations.

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Most popular questions from this chapter

Formic acid, \(\mathrm{HCHO}_{2},\) is a convenient source of small quantities of carbon monoxide. When warmed with sulfuric acid, formic acid decomposes to give CO gas. $$ \mathrm{HCHO}_{2}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}(g) $$ If \(3.85 \mathrm{~L}\) of carbon monoxide was collected over water at \(25^{\circ} \mathrm{C}\) and \(689 \mathrm{mmHg}\), how many grams of formic acid were consumed?

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