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e-lub [12.9K]
3 years ago
12

What volume would 75.0g of oxygen gas occupy

Chemistry
1 answer:
seropon [69]3 years ago
7 0

Answer:

Explanation: It is already known that 1 mole of the gas( or 32g of O2) is equivalent to 22.4 Litres of the oxygen gas. So, 8g is equivalent to = (22.4/32) × 8 = 5.6 L of the gas.

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What is Catenation in chemistry?​
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What is the volume of ammonia produced at 243 K at a pressure of 1.38 atm by the unbalanced reaction on the left if 5740 moles o
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Answer:

49671 L is the produced volume of ammonia

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We have 5740 moles of hydrogen, so we have enough hydrogen. This is the excess reagent, so the limiting is the N₂

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V = (n . R . T) /P → V = (3440 mol . 0.082 L.atm/mol.K . 243K) / 1.38 atm

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We confirm that the nitrogen was the limiting reactant

3 moles of H₂ need 1 mol of nitrogen to react

Therefore, 5740 moles of H₂ will react with (5740 . 1) /3 = 1913 moles of N₂

It was ok to say, that N₂ was the limiting reactant because we need 1913 moles in the reaction, and we only have 1720 moles

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