Can someone please help me with this stoichiometry problem?

[14cwelch]

A beaker containing 3.400g of Hexane (C₆H₈) gas is combusted inside a cubic sealed chamber which measures 13.78 inches on each edge. The chamber's interior temperature was 75.00 degrees Fahrenheit before the combustion reaction. The container had an interior pressure of 1.000atm prior to the combustion. The air in the chamber was 20.000% oxygen by volume prior to the combustion reaction. After combustion, the interior temperature rose to 91.00 degrees Fahrenheit.

1. Determine the limiting reactant in the combustion reaction. How many leftover grams of the excess product remain? (should equal 0.3808g excess)

2. Determine the final pressure in the chamber after the combustion reaction in mmHg. Remember to account for the Oxygen used in the combustion and the gaseous products generated in the combustion. Assume that any water formed as a reaction product is vapor. (should equal 820.4mmHg)

Useful Facts: 1in = 2.54cm; 1000cm^3 = 1.000L; C= 5/9(F-32); PV=nRT; R=.08206L*atm/mol*K; 760mmHg = 1.0atm

I already figured out how to do the first one but the second question is confusing me so if someone could show me the work they used or explain to me how to answer it, that would be great.

[GrahamKemp]

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P1 / P0 = n1 T1 / n0 T0

.... volume and ideal gas constant are constant.

Known: initial pressure

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P0

, initial and final temperatures

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T0

and

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T1

. Volume of Chamber,

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V

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P0

= 1.000 atm

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 T0 = ( 75.00 - 32.00 ) 5 / 9 + 273.15 K = 297.04 K 

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 T1 = ( 91.00 - 32.00 ) 5 / 9 + 273.15 K = 305.93 K 

From this: Initial amount of gas:

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n0 = P0 V / R T0 

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n0 = ( 1.000 atm 42.88 L ) / ( 0.8206 atm L/mol K * 297.04 K ) = 0.1759 mol

NB 20% of this is the initial amount of oxygen

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 0.03518 mol

The initial amount of hexane is:

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 3.400g / 80.12826 g/mol = 0.04243 mol

Find: final amount of gas,

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n1

, knowing:
Equation:

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C6H8 (l) + 5 O2 (g) = 6 CO (g) + 4 H2O (g)

Mole amount of hexane consumed:

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x = 0.03518/5 = 0.007037 mol (=0.5638 g)

Thus:

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n1 = n0 + 5x

Thus by algebra:

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P1 = (P0 / T0 + 5x R / V) T1

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P1 = ( 1.000 atm / 297.04 K + 0.03518 mol * 0.8206 atm L/mol K / 42.88 L) * 305.93 K = 1.236 atm

(NB: The mole amount, x, is determined by your answer to the first half. If hexane is the limiting reactant: x is all the hexane; if oxygen is the limiting reactant: x is one fifth of all the oxygen.)