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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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Most popular questions from this chapter

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(\mathrm{~g})\), ammonia, \(\mathrm{NH}_{3}(\mathrm{~g})\), and oxygen, \(\mathrm{O}_{2}(\mathrm{~g})\), at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of \(\mathrm{HCN}(g)\) can be obtained from the reaction of \(20.0 \mathrm{~L} \mathrm{CH}_{4}(g), 20.0 \mathrm{~L} \mathrm{NH}_{3}(g)\), and \(20.0 \mathrm{~L} \mathrm{O}_{2}(\mathrm{~g})\) ? The volumes of all gases are measured at the same temperature and pressure.

Equal moles of sulfur dioxide gas and oxygen gas are mixed in a flexible reaction vessel and then sparked to initiate the formation of gaseous sulfur trioxide. Assuming that the reaction goes to completion, what is the ratio of the final volume of the gas mixture to the initial volume of the gas mixture if both volumes are measured at the same temperature and pressure?

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) $$ What volume of pure \(\mathrm{O}_{2}(\mathrm{~g})\), collected at \(27^{\circ} \mathrm{C}\) and 746 torr, would be generated by decomposition of \(125 \mathrm{~g}\) of a \(50.0 \%\) by mass hydrogen peroxide solution? Ignore any water vapor that may be present.

An organic compound containing only C, H, and N yields the following data. i. Complete combustion of \(35.0 \mathrm{mg}\) of the compound produced \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\). ii. A \(65.2\) -mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129), giving \(35.6 \mathrm{~mL}\) of dry \(\mathrm{N}_{2}\) at 740 . torr and \(25^{\circ} \mathrm{C}\). iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{~mL} / \mathrm{min}\). The effusion rate of argon gas, under identical conditions, is \(26.4 \mathrm{~mL} / \mathrm{min}\). What is the molecular formula of the compound?

A bicycle tire is filled with air to a pressure of 75 psi at a temperature of \(19^{\circ} \mathrm{C}\). Riding the bike on asphalt on a hot day increases the temperature of the tire to \(58^{\circ} \mathrm{C}\). The volume of the tire increases by \(4.0 \%\). What is the new pressure in the bicycle tire?
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