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Chapter 5: Problem 52

Calculate the mass in grams of hydrogen chloride produced when \(5.6 \mathrm{~L}\) of molecular hydrogen measured at STP react with an excess of molecular chlorine gas.

Short Answer

Expert verified
18.25 g of HCl is produced.

Step by step solution

01

Understand the Reaction

The chemical reaction between molecular hydrogen \((H_2)\) and molecular chlorine \((Cl_2)\) produces hydrogen chloride \((HCl)\). The balanced chemical equation for this reaction is: \[H_2(g) + Cl_2(g) \rightarrow 2HCl(g)\]. This tells us that one molecule of \(H_2\) reacts with one molecule of \(Cl_2\) to produce two molecules of \(HCl\).
02

Use the Ideal Gas Law at STP

At standard temperature and pressure (STP), 1 mole of any gas occupies \(22.4\mathrm{~L}\). Given \(5.6\mathrm{~L}\) of \(H_2\), we calculate the number of moles of \(H_2\) by using the relationship: \[\text{Number of moles} = \frac{\text{Volume at STP}}{22.4 \mathrm{~L/mol}}\]. Thus, the moles of \(H_2\) are \(\frac{5.6\mathrm{~L}}{22.4 \mathrm{~L/mol}} = 0.25 \text{ moles}\).
03

Determine Moles of HCl Produced

From the balanced equation, 1 mole of \(H_2\) produces 2 moles of \(HCl\). Therefore, for \(0.25\) moles of \(H_2\), the moles of \(HCl\) produced are \(2 \times 0.25 = 0.5 \text{ moles}\).
04

Calculate Mass of HCl Produced

The molar mass of \(HCl\) is \(1 + 35.5 = 36.5 \text{ g/mol}\). To find the mass produced, multiply the moles of \(HCl\) by its molar mass: \(0.5 \text{ moles} \times 36.5 \text{ g/mol} = 18.25 \text{ g}\).

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Key Concepts

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

Chemical Reactions
In chemistry, a chemical reaction occurs when substances interact to form one or more new substances. A classic example is the reaction between molecular hydrogen \((H_2)\) and molecular chlorine \((Cl_2)\). This produces hydrogen chloride \((HCl)\). The balanced equation for this reaction is \[H_2(g) + Cl_2(g) \rightarrow 2HCl(g)\]. It tells us about the proportions of reactants and products involved, ensuring mass and atoms are conserved.
During such reactions:
  • Reactants rearrange their atoms to form new chemical structures.
  • The law of conservation of mass applies, meaning the mass of reactants equals the mass of products.
Understanding chemical equations helps predict the amount of product formed from given reactants.
Molar Mass
Molar mass is a crucial concept in chemistry as it relates to the mass of one mole of a substance. For any element or compound, it's expressed in grams per mole (g/mol). This allows us to convert between moles and grams, an essential step in stoichiometry.
To calculate the molar mass:
  • Sum the atomic masses of each atom in the molecule.
  • Units are typically in atomic mass units \(u\), which converts directly to \(g/mol\).
For example, the molar mass of hydrogen chloride \(HCl\) is computed as follows:
  • Hydrogen's atomic mass = 1 u
  • Chlorine's atomic mass = 35.5 u
  • Total molar mass = 1 + 35.5 = 36.5 g/mol
This helps in determining the amount in grams needed or produced in reactions.
Ideal Gas Law
The ideal gas law is an equation of state that describes how gases behave under different conditions of temperature, volume, and pressure. The formula is \(PV = nRT\), where:
  • \(P\) is the pressure in atmospheres (atm)
  • \(V\) is the volume in liters (L)
  • \(n\) is the number of moles
  • \(R\) is the universal gas constant (0.0821 L atm/mol K)
  • \(T\) is the temperature in Kelvin (K)
At Standard Temperature and Pressure (STP), 1 mole of gas occupies \(22.4\) liters. In our example, by knowing the volume of hydrogen gas at STP, we could find its mole quantity using \(\frac{V}{22.4} = n\). This relationship is foundational for predicting gas behavior in reactions.
STP Conditions
STP stands for standard temperature and pressure, providing a set uniform condition to measure gases. It simplifies the stoichiometry, calculation, and comparison across different scenarios.
STP conditions are:
  • Temperature: 0 degrees Celsius, or 273.15 Kelvin
  • Pressure: 1 atm
  • Volume: 22.4 L for gases per mole
Utilizing these constants, chemists effortlessly convert volumes to moles when dealing with gases at STP. In our problem, this uniformity allowed us to determine that \(5.6 \text{ L}\) of hydrogen gas equates to 0.25 moles. This standardization plays a crucial role in simplifying complex calculations across sessions and applications.

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Most popular questions from this chapter

Given that 6.9 moles of carbon monoxide gas are present in a container of volume \(30.4 \mathrm{~L},\) what is the pressure of the gas (in atm) if the temperature is \(62^{\circ} \mathrm{C} ?\)

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The pressure of \(6.0 \mathrm{~L}\) of an ideal gas in a flexible container is decreased to one-third of its original pressure, and its absolute temperature is decreased by one-half. What is the final volume of the gas?

An ideal gas originally at \(0.85 \mathrm{~atm}\) and \(66^{\circ} \mathrm{C}\) was allowed to expand until its final volume, pressure, and temperature were \(94 \mathrm{~mL}, 0.60 \mathrm{~atm},\) and \(45^{\circ} \mathrm{C}\), respectively. What was its initial volume?

Under constant-pressure conditions a sample of hydrogen gas initially at \(88^{\circ} \mathrm{C}\) and \(9.6 \mathrm{~L}\) is cooled until its final volume is \(3.4 \mathrm{~L}\). What is its final temperature?
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