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Chapter 5: Problem 45

An 11.2-L sample of gas is determined to contain \(0.50\) mole of \(\mathrm{N}_{2}\). At the same temperature and pressure, how many moles of gas would there be in a 20.-L sample?

Short Answer

Expert verified
The number of moles in the 20 L sample can be determined using the proportionality relationship between volume and moles of a gas. Given a 11.2 L sample containing 0.50 moles of N2, we can calculate the number of moles in the 20 L sample as: \[n_2 = \frac{n_1}{V_1} \times V_2\] By substituting the given values and performing the calculations, we find that the 20 L sample contains approximately 0.89 moles of gas (rounded to 2 decimal places).

Step by step solution

01

Identify the proportionality relationship between volume and moles of a gas

Under the same temperature and pressure conditions, the volume of a gas is directly proportional to its moles. We can use this relationship to find the number of moles in the different volume. Let's denote the given values as follows: - \(V_1 = 11.2 \,\text{L}\) (volume of the first sample) - \(n_1 = 0.50\,\text{mole}\) (moles of N2 in the first sample) - \(V_2 = 20\,\text{L}\) (volume of the second sample) The relationship between the two samples can be written as: \[\frac{n_1}{V_1} = \frac{n_2}{V_2}\] Where \(n_2\) is the number of moles in the 20 L sample, which is the value we need to find.
02

Rearrange the equation for the unknown value

We need to isolate \(n_2\) in the proportionality equation above. Rearrange the equation to solve for the unknown value, which is \(n_2\): \[n_2 = \frac{n_1}{V_1} \times V_2\]
03

Substitute the given values in the equation

Now that the equation is ready, substitute the given values for the volume and moles of the first sample and the volume of the second sample: \[n_2 = \frac{0.50\,\text{mole}}{11.2\,\text{L}} \times 20\,\text{L}\]
04

Calculate the number of moles in the 20 L sample

Perform the calculations to find the value of \(n_2\): \[n_2 = \frac{0.50\,\text{mole}}{11.2\,\text{L}} \times 20\,\text{L} = 0.8929\,\text{mole}\] The number of moles in the 20 L sample is approximately 0.89 moles (rounded to 2 decimal places).

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Key Concepts

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

Volume-to-mole relationship
The volume-to-mole relationship is a fundamental concept in the study of gases. It is derived from the Ideal Gas Law, which states that for a given amount of gas at constant temperature and pressure, the volume of the gas is directly proportional to the number of moles. This means that if you know the volume of a gas, you can determine its moles, and vice versa.
  • If the volume of gas increases, the number of moles increases as long as temperature and pressure remain constant.
  • Conversely, if the volume decreases, the number of moles decreases under the same conditions.
This relationship is powerful because it allows us to predict how changes in volume affect the quantity of gas. This prediction is based solely on proportional changes between volume and moles.
Proportionality in gases
Proportionality in gases focuses on how various properties such as volume, pressure, and temperature interact with the quantity of gas. For gases, under conditions where temperature and pressure are constant, volume is directly proportional to the number of moles.
To understand this better, think of gases as collections of tiny particles. When you have more particles (or moles), they take up more space, thus increasing volume. This proportional relationship can be expressed mathematically as:
\[\frac{n_1}{V_1} = \frac{n_2}{V_2}\]
  • This equation shows that the ratio of moles to volume is consistent across different samples of a gas.
  • This invariant ratio enables scientists and students to find unknown quantities based on changes in volume or moles.
Understanding this proportionality allows for practical calculations in laboratory settings or everyday situations involving gases.
Moles calculation
Moles calculation is a critical step in solving problems involving gases. By using known quantities and the relationships defined by the Ideal Gas Law, you can solve for unknowns effectively. In our context, we are applying the volume-to-mole proportionality.
Here is how you typically approach a moles calculation:1. **Identify known values:** Start by determining what you already know 鈥� such as initial volume, initial moles, and new volume. 2. **Set up the relationship:** Use the proportional relationship \(\frac{n_1}{V_1} = \frac{n_2}{V_2}\) to set up your calculation.3. **Solve for the unknown:** Rearrange the formula to solve for the unknown quantity, which is usually either moles or volume.
  • Make sure each unit of measure is in consistent units to avoid errors.
  • Plug in the numbers correctly to find the solution.
This approach to moles calculation helps demystify seemingly complex problems and provides a clear pathway from known to unknown values.

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Most popular questions from this chapter

Draw a qualitative graph to show how the first property varies with the second in each of the following (assume 1 mole of an ideal gas and \(T\) in kelvin). a. \(P V\) versus \(V\) with constant \(T\) b. \(P\) versus \(T\) with constant \(V\) c. \(T\) versus \(V\) with constant \(P\) d. \(P\) versus \(V\) with constant \(T\) e. \(P\) versus \(1 / V\) with constant \(T\) f. \(P V / T\) versus \(P\)

Silane, \(\mathrm{SiH}_{4}\), is the silicon analogue of methane, \(\mathrm{CH}_{4}\). It is prepared industrially according to the following equations: $$ \begin{aligned} \mathrm{Si}(s)+3 \mathrm{HCl}(g) & \longrightarrow \operatorname{HSiCl}_{3}(l)+\mathrm{H}_{2}(g) \\ 4 \mathrm{HSiCl}_{3}(l) & \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l) \end{aligned} $$ a. If \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}(d=1.34 \mathrm{~g} / \mathrm{mL})\) is isolated when \(15.0 \mathrm{~L}\) \(\mathrm{HCl}\) at \(10.0 \mathrm{~atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3}\) ? b. When \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at \(10.0\) atm and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is \(93.1 \%\) ?

Consider the following chemical equation. $$ 2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ If \(25.0 \mathrm{~mL} \mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?

The total mass that can be lifted by a balloon is given by the difference between the mass of air displaced by the balloon and the mass of the gas inside the balloon. Consider a hot-air balloon that approximates a sphere \(5.00 \mathrm{~m}\) in diameter and contains air heated to \(65^{\circ} \mathrm{C}\). The surrounding air temperature is \(21^{\circ} \mathrm{C}\). The pressure in the balloon is equal to the atmospheric pressure, which is 745 torr. a. What total mass can the balloon lift? Assume that the average molar mass of air is \(29.0 \mathrm{~g} / \mathrm{mol}\). (Hint: Heated air is less dense than cool air.) b. If the balloon is filled with enough helium at \(21^{\circ} \mathrm{C}\) and 745 torr to achieve the same volume as in part a, what total mass can the balloon lift? c. What mass could the hot-air balloon in part a lift if it were on the ground in Denver, Colorado, where a typical atmospheric pressure is 630 . torr?

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of 200. L/min at \(1.50 \mathrm{~atm}\) and ambient temperature. Air is added to the chamber at \(1.00 \mathrm{~atm}\) and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(g)\), three times as much oxygen as is necessary is reacted. Assuming air is 21 mole percent \(\mathrm{O}_{2}\) and \(79 \mathrm{~mole}\) percent \(\mathrm{N}_{2}\), calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that \(95.0 \%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\). The remainder was present as carbon in \(\mathrm{CO}\). Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O} .\) Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.
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