What will be the pressure exerted (in pascal) by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9dm3(3-cube) flask at 300K?

Asked by  | 2nd Dec, 2012, 11:28: PM

Expert Answer:

For methane, CH4 (molar mass = 16 g/mol)

moles = mass / molar mass
= 3.2 g / 16 g/mol
= 0.20 mol


for carbon dioxide (molar mass = 44 g/mol)

moles = mass / molar mass
= 4.4 g / 44 g/mol
= 0.10 mol

So, altogether we have 0.30 mol of gas
 
Step 2: Now we use the ideal gas equation to find the pressure.

PV = nRT
so P = nRT / V

n = 0.30 mol
R = 8.31 kPa.L/mol.K
T = 27°C = 300 K
V = 9.0 L

P = (0.30 mol)(8.31 kPa.L/mol.K)(300 K) / (9.0 L)
= 83 kPa (2 significant digits)

Answered by  | 5th Dec, 2012, 10:19: AM

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