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Ratling [72]
3 years ago
8

Determine the volume occupied by 2.5 mol of a gas at 18 °C if the pressure is 81.8 kPa?

Chemistry
1 answer:
mars1129 [50]3 years ago
3 0

Answer:

74 litre

Explanation:

using ideal gas eqation PV=nRT

here P(pressure)=81.8 kPa =81.8×10^3 Pa

moles=2.5

temperature=273.15+18=291.15K

Gas constant R=8.314m^3-Pa/K-mol

now, V=nRT/P = 8.314×2.5×291.5/81.8×10^3 ≈74litre

✌️;)

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<u>Answer:</u> The total volume of the gaseous products is 1044.29 L

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By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water vapor

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Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.

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At STP, 1 mole of gas occupies 22.4 L of volume

Given,

Volume of ammonia reacted = 0.720 L

The combustion reaction is shown as,

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(4 \times  22.4) L of ammonia reacts with (5 \times  22.4) L of oxygen gas.

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\dfrac{ (5 \times  22.4)}{(4 \times  22.4)} \times 0.720 = 0.9 \;\rm L

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