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Snezhnost [94]
3 years ago
9

Consider the reaction of gaseous hydrogen with gaseous oxygen to produce gaseous water. Given that the first picture represents

12 liters of hydrogen molecules, determine the volumes represented by each of the other images. Assume all images are at the same constant temperature and pressure. What is the H_{2}O molecule volume

Chemistry
1 answer:
Bogdan [553]3 years ago
6 0

The question is incomplete. There's missing the image, which is shown below.

Answer:

Volume of O₂ = 6 L, volume of mixture: 18 L, volume of H₂O = 12 L, molecule volume of H₂O = 0.667 molecule/L

Explanation:

The reaction between hydrogen gas and oxygen gas to form water is:

2H₂(g) + O₂(g) → 2H₂O(g)

So, for 1 mol of O₂ is necessary 2 moles of H₂ form 2 moles of H₂O. As the images below there's 8 molecules of H₂, 4 molecules of O₂, 12 molecules in the mixture, and 8 molecules of H₂O. Thus, there are stoichiometric values.

All the images are at the same temperature and pressure, so, by the ideal gas law:

PV= nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The number of moles and molecules are related, so let's substitute it in the equation. For the H₂:

P*12 = 8*RT

RT/P = 12/8 = 1.5

Thus, for O₂:

PV= nRT

V = n*(RT/P)

V = 4*1.5 = 6 L

For the mixture:

V = 12*1.5 = 18 L

For H₂O:

V = 8*1.5 = 12 L

The molecule volume is the number of molecules divided by the volume they occupy, thus for water: 8/12 = 0.667 molecules/L

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A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t
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Answer:

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Explanation:

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where P is pressure

v is Volume/mol

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[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K  .  296.9K

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0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

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