Oxygen gas reacts with powdered aluminum according to the reaction: 4 Al(s) + 3 O 2 (g) → 2 Al 2 O 3 (s) What volume of O 2 gas (in L), measured at 782 mmHg and 25 °C, completely reacts with 53.2 g Al?

Hey everyone. So you have chlorine from floor to floor methane ozone, we have a series of reactions below. and at 32 mm of Mercury, 288 Kelvin. We assume that all the chlorine from 16.5 g of methane undergoes 15 cycles. Or as calculate the total volume of ozone that is consumed in ozone is 03. There for one cycle we need to convert from grams of chlorophyll and methane thomas of ozone Start giving 16.5 g of course truffle or methane. And in one mole of course our floor methane of the Mueller mass Which is 12.011 plus three, 18. plus 35 .453 Mr. give us 104 0.46 gramps. And for one mode of course to florida, methane, We have two moles of ozone in one cycle And give us 0.3159. Malls of ozone dynamic is the ideal gas law equation. To find the volume, it's gonna be P. V. Equals Nrt our pressure. It's 32 millimeters of mercury. We're looking for the volume number of moles, 0.3159. And our constant Can be 0.08 - one. Later. Sounds atmosphere about about moles. I'm kelvin for the temperature Gonna have 288 Melvin. And now we need to convert the pressure from millimeters of mercury to atmosphere 32 kilometers of mercury. We have 760 millimeters of mercury in one atmosphere, You need to give us 0.04 21 atmosphere. You know, if you just plug in the values to the equation to get 0.04 atmosphere times of volume, It's going to be equal to 0.31 59 balls Time 0.08 21. Their sounds atmosphere about about malls. I'm Calvin. I'm Sergeant 88 Calvin committed by both sides by 0.04 atmosphere. We're gonna get the volume. It's gonna be 177 0.4 leaders. But that is in one cycle, We need to determine the volume in 15 cycles. They're going to take 177.4 leaders in one cycle Most about 15 cycles. And this gave us 2661 leaders, Which is 2.7 Times 10 to the 3rd Leaders. Thanks for watching my video and I hope it was helpful.

Ch.6 - Gases

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Tro 6th Edition

Ch.6 - Gases

Problem 82

Chapter 6, Problem 82

Oxygen gas reacts with powdered aluminum according to the reaction: 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s) What volume of O₂ gas (in L), measured at 782 mmHg and 25 °C, completely reacts with 53.2 g Al?

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Calculate the moles of aluminum (Al) using its molar mass. The molar mass of Al is approximately 26.98 g/mol.

Use the stoichiometry of the balanced chemical equation to find the moles of oxygen gas (O2) needed. According to the equation, 4 moles of Al react with 3 moles of O2.

Convert the moles of O2 to volume using the ideal gas law equation: PV = nRT. First, convert the pressure from mmHg to atm (1 atm = 760 mmHg).

Convert the temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature.

Substitute the values into the ideal gas law equation to solve for the volume (V) of O2 gas. Use R = 0.0821 L·atm/mol·K as the ideal gas constant.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced equation. It allows us to determine the proportions of substances involved in a reaction. In this case, the balanced equation shows that 4 moles of aluminum react with 3 moles of oxygen gas, which is essential for calculating the amount of O2 needed for the given mass of Al.

Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas through the equation PV = nRT. This law is crucial for converting the amount of oxygen gas from moles to volume under specific conditions of temperature and pressure. Understanding how to manipulate this equation will help in finding the volume of O2 required for the reaction.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is essential for converting grams of a substance to moles, which is necessary for stoichiometric calculations. For aluminum, the molar mass is approximately 26.98 g/mol, and knowing this allows us to determine how many moles of Al are present in 53.2 g, which is the first step in solving the problem.

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Oxygen gas reacts with powdered aluminum according to the reaction: 4 Al(s) + 3 O2(g) → 2 Al2O3(s) What volume of O2 gas (in L), measured at 782 mmHg and 25 °C, completely reacts with 53.2 g Al?

Verified video answer for a similar problem:

Key Concepts

Stoichiometry

Ideal Gas Law

Molar Mass

Oxygen gas reacts with powdered aluminum according to the reaction: 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s) What volume of O₂ gas (in L), measured at 782 mmHg and 25 °C, completely reacts with 53.2 g Al?