Chem                    Quiz #8

 

  1. (15 pt.)  Under intense heat, rust (Fe2O3) can be reduced back to iron metal with the release of oxygen gas.  If 250 g of rust were placed in the bottom of a 2 L evacuated flask and heated to 850 ºC, what would the final pressure be inside the flask?  What would the pressure be after the flask was allowed to cool down to room temperature (25ºC)?

 

[If you’re not sure what the reaction is, you can buy a balanced equation for 3 points.]

Molar mass of O = 16 g/mol

Molar mass of Fe = 55.845 g/mol

 

2 Fe2O3 (s) Y 4 Fe (s) + 3 O2 (g)

 

250 g Fe2O3 * ( 1 mol Fe2O3/159.69 g Fe2O3) = 1.566 mol Fe2O3

 

1.566  mol Fe2O3 * (3 mol O2/2 mol Fe2O3) = 2.35 mol O2

 

O2 is the only gas and, therefore the only source of pressure

 

PV=nRT

850 C + 273.15 = 1123 K

 

P = nRT/V = (2.35 mol) (0.082056 Latm/mol K) (1123 K)/2 L

P = 108.3 atm

 

P = nRT/V = (2.35 mol) (0.082056 Latm/mol K) (298 K)/2 L

P = 28.73 atm

 

 

 

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