Calculate the pressure inside the vessel at 400°C.?
Ferrous oxalate, FeC2O4, will decompose on heating:
FeC2O4 (s) ® FeO (s) + CO2 (g) + CO (g)
A 1.25-gram sample of FeC2O4 is added to an evacuated 2.00-liter steel vessel.
The vessel is heated to 400°C, at which point all the FeC2O4 is decomposed.Calculate the pressure inside the vessel at 400°C.
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Molar mass FeC2O4 = 143.8645 g/mol
1.25g FeC2O4 = 8.689*10^-3 molFeC2O4
From the balanced equation,
1mol FeC2O4 will produce 1mol CO2 and 1mol CO
8.689*10^-3 mol FeC2O4 will produce 8.689*10^-3 mol CO2 and 8.689*10^-3 mol CO
In total you will have 2*(8.689*10^-3) mol = 0.0174 mol mixed gases.
Use gas equation:
PV = nRT
P * 2 = 0.0174*0.082057*673
P = 0.48 atm
The pressure will be 0.48 atm
I will try to explain but do not copy and paste, try your best to understand the method on how to apply Gas Law.
Use Gas Law PV/T = constant coupled with the facts that 1 mole of an ideal gas occupies 22.414 at STP.
Method
1). Determine the molar mass ofr all components involved in the chemical reaction
2). Do mass balance
3). Use Gas Law to calculate the pressure
Solution
1) Molar mass: FeC2O4 = 143.85 , CO2 = 44, CO = 28
2). Mass balance
I mole of FeC2O4 produces 1 mole of CO2 and 1 Mole of CO
In 1.25 g FeC2O4 = 1.25 / 143.85 = 0.0087
Thus, the total mole of gas mixture produced = 2 x 0.0087 = 0.0174 mole
3). Gas Law can be written as
P2V2/T2 = Constant = P1V1/T1
At STP, V1 = 22.414 L, P1 = 1 atm, T1 = 273.15 K
Given that V2 = 2 L and T2 = 400 +273.15 = 673.15 K
Thus, P2 = [1] x [22.414 / 2] x [673.15 / 273.15] = 27.6185 atm
The above pressure is occupied by 1 mole of gas mixture.
So for 0.0174 mole of gas,
P2= 0.0174 x 27.6185 = 0.48 atm