Question

A 2.0 L container at 25°C contains 1.25 mol of oxygen and 3.3 mol of carbon(s). (a) What is the initial pressure in the flask? (b) If carbon and oxygen react as completely as possible to form CO, what will be the final pressure in the container?​

Answer 1

(i) The container contains 1.25 moles of O2 (gas) and 3.2 moles of carbon (solid).Since, only O2, is gaseous and carbon will not exert any pressure. So, n = 1.25 mol V = 2.0 L T = 25 + 273 = 298 K According to ideal gas equation, PV = nRT ∴∴ P = nRTVnRTV = 1.25mol×0.0821Latmmol−1K−1×298K2.0L1.25mol×0.0821Latmmol−1K−1×298K2.0L = 15.3 atm (ii) ∴∴ 1212 mole of O2 gives = 1 mol of co.

= 15.3 atm (ii) ∴∴ 1212 mole of O2 gives = 1 mol of CO ∴∴ 1.25 moles of O2 will give = 1×1.25121×1.2512 = 1 x 1.25 × 2 = 2.50 mol ∴∴ Final pressure P = nRTVnRTV = 2.50mol×0.0821Latmmol−1K−1×298K2.0L2.50mol×0.0821Latmmol−1K−1×298K2.0L = 30.6 Atm

Answer 2

Answer:

1) 15.3atm

2) 30.6atm

Explanation:

i) The container contains 1.25 moles of O2 (gas) and 3.2 moles of carbon (solid).

Since, only O2, is gaseous and carbon will not exert any pressure.

So, n = 1.25 mol

V = 2.0 L

T = 25 + 273 = 298 K

According to ideal gas equation, PV = nRT

∴ P = nRT/V

=(1.25×0.0821×298)/2.0

=15.3 atm

ii)∴  1/2 mole of O2 gives = 1 mol of CO

=> 1.25 moles of O2 will give = 2.50 mol

∴ Final pressure P = nRT/V

=(2.50×0.0821×298)/2.0

= 30.6 atm

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