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weeeeeb [17]
3 years ago
5

8). How many moles of oxygen gas must be placed in a 35.0 L container in order

Chemistry
1 answer:
klemol [59]3 years ago
5 0

Answer:

4.1 moles

Explanation:

Applying

PV = nRT................ equation 1

Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = Temperature.

make n the subject  of the equation

n = PV/RT.............. Equation 2

From the question,

Given: V = 35 L , P = 2.8 atm, T = 15 °C  = (15+273) = 288 K, R = 0.083 L.atm/K.mol

Substitute these values into equation 2

n = (35×2.8)/(0.083×288)

n = 4.1 moles

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Julli [10]

Answer:

2.5 moles of Al

Explanation:

We'll begin by calculating the number of mole in 127 g of Al₂O₃. This can be obtained as follow:

Mass of Al₂O₃ = 127 g

Molar mass of Al₂O₃ = 101.961 g/mol

Mole of Al₂O₃ =?

Mole = mass / molar mass

Mole of Al₂O₃ = 127 / 101.961

Mole of Al₂O₃ = 1.25 mole

Finally, we shall determine the number of mole of Al that reacted. This can be obtained as follow:

4Al + 3O₂ —> 2Al₂O₃

From the balanced equation above,

4 moles of Al reacted to produce 2 moles of Al₂O₃.

Therefore, Xmol of Al will react to produce 1.25 moles of Al₂O₃ i.e

Xmol of Al = (1.25 × 4)/2

Xmol of Al = 2.5 moles.

Thus, 2.5 moles of Al is needed for the reaction.

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