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Calculate the density of hydrogen sulfide gas, \(\mathrm{H}_{2} \mathrm{~S},\) at \(49^{\circ} \mathrm{C}\) and \(967 \mathrm{mmHg} .\) Obtain the density in grams per liter.

Short Answer

Expert verified
The density of \(\text{H}_2\text{S}\) is approximately 1.639 g/L at the given conditions.

Step by step solution

01

Identify the Gaseous Equation

To find the density of a gas, we first need to use the Ideal Gas Law, which is stated as \( PV = nRT \). We can substitute \( n \) (the number of moles) using \( n = \frac{m}{M} \), where \( m \) is the mass, and \( M \) is the molar mass. The equation becomes \( PV = \frac{m}{M}RT \).
02

Convert Units for Ease of Calculation

Convert the temperature from Celsius to Kelvin using the formula: \( T = 49 + 273.15 = 322.15 \text{ K} \). Convert the pressure from mmHg to atm: \( 967 \text{ mmHg} \times \frac{1\text{ atm}}{760\text{ mmHg}} = 1.272\text{ atm} \).
03

Rearrange the Formula for Density

Rearrange the Ideal Gas Law for density \( d = \frac{m}{V} \). Substitute \( m = PMV/RT \). So the equation becomes \( d = \frac{PM}{RT} \).
04

Calculate the Molar Mass of \(\text{H}_2\text{S}\)

Hydrogen sulfide \(\text{H}_2\text{S} \) consists of 2 hydrogen atoms and 1 sulfur atom. The molar mass is calculated as:- Hydrogen: \(2 \times 1.008 \text{ g/mol} = 2.016 \text{ g/mol}\)- Sulfur: \(32.065 \text{ g/mol}\)Adding these gives us \(M = 34.081 \text{ g/mol}\).
05

Insert Values into the Density Equation

Now substitute into the density formula: \[ d = \frac{(1.272 \text{ atm})(34.081 \text{ g/mol})}{(0.0821 \text{ L atm/mol K})(322.15 \text{ K})}\] Calculate: \[ d = \frac{43.330632 \text{ g/L atm}}{26.444915 \text{ L atm/mol K}} \approx 1.639 \text{ g/L}\]
06

Conclusion and Verification

The calculated density of hydrogen sulfide gas at \(49^{\circ} \, \text{C}\) and \(967 \, \text{mmHg}\) is approximately \(1.639 \, \text{g/L}\). Double-check by ensuring units cancel out correctly, confirming that the calculations are accurate.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation that mathematically describes the behavior of an ideal gas. It combines several gas laws, including Boyle's, Charles's, and Avogadro's laws, into one equation:
  • \(PV = nRT\)
where
  • \(P\) is the pressure of the gas,
  • \(V\) is the volume,
  • \(n\) is the number of moles,
  • \(R\) is the ideal gas constant (0.0821 L atm/mol K),
  • and \(T\) is the temperature in Kelvin.
The equation expresses how the pressure and volume of a gas change as a function of its temperature and quantity. Additionally, it's really helpful for relating the gas's properties to one another.
By substituting the number of moles \(n\) with \(\frac{m}{M}\), where \(m\) is the mass, and \(M\) is the molar mass, the equation can be reorganized to solve for density, which brings us closer to calculating the density of a gas.
Hydrogen Sulfide
Hydrogen sulfide (H2S) is a colorless gas with the characteristic foul odor of rotten eggs. It is commonly found in nature and can be produced during the decomposition of organic substances. Hydrogen sulfide is slightly denser than air and can be toxic in high concentrations.
  • It's composed of two hydrogen atoms and one sulfur atom.
  • In practical applications, this gas is crucial in industries such as mining and chemical manufacturing.
  • Despite its smell, hydrogen sulfide plays an important role in various biological processes.
In our exercise, we're looking into the density of hydrogen sulfide under certain conditions. To do this, we utilize its molecular composition and use it in our calculations to find its density using the ideal gas law.
Molar Mass
Molar mass is the mass of one mole of a particular substance, usually expressed in grams per mole (g/mol). For calculation purposes, it's crucial to accurately determine the molar mass of any gaseous compound.
In our example, we need to find the molar mass of hydrogen sulfide, \(\mathrm{H}_2\mathrm{S}\).
  • Each hydrogen atom has an atomic mass of approximately 1.008 g/mol.
  • There are two hydrogen atoms, so altogether they contribute about 2.016 g/mol.
  • The sulfur atom has an atomic mass of 32.065 g/mol.
Adding these elements together gives us the total molar mass of hydrogen sulfide: \[M = 2.016 \, \text{g/mol} + 32.065 \, \text{g/mol} = 34.081 \, \text{g/mol}\]This value is critical when substituting into formulas involving the ideal gas law, particularly to find density.
Unit Conversion
In scientific calculations, especially those involving gases, accurately converting units is essential to obtaining a correct result. Let's look at the necessary conversions that apply to our exercise.
  • Temperature: Always convert Celsius to Kelvin by adding 273.15 because the gas law equations require temperature in absolute terms.
  • Pressure: Convert pressure from millimeters of mercury (mmHg) to atmospheres (atm) using the conversion factor \(1 \text{ atm} = 760 \text{ mmHg}\), since \(R\)'s value is based on atm.
For instance, our exercise involves converting
  • \(49 \degree \text{C} + 273.15 = 322.15 \text{ K}\)
  • and \(967 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 1.272 \text{ atm}\)
Such conversions are crucial to apply the ideal gas law correctly, ensuring that the results, like the calculated density, are accurate.

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Most popular questions from this chapter

A \(24.9-\mathrm{mL}\) volume of hydrochloric acid reacts completely with \(55.0 \mathrm{~mL}\) of aqueous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). The reaction is \(2 \mathrm{HCl}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(t)+2 \mathrm{NaCl}(a q)\) The volume of \(\mathrm{CO}_{2}\) formed is \(141 \mathrm{~mL}\) at \(27^{\circ} \mathrm{C}\) and \(727 \mathrm{mmHg}\). What is the molarity of the HCl solution?

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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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