Pentane gas (C_3 H_12) reacts with molecular oxygen to form carbon dioxide and g

Question

Pentane gas (C_3 H_12) reacts with molecular oxygen to form carbon dioxide and gaseous water. Balance chemical reaction below for this process. Note that this equation applies to all parts of problem 6. C_3 H_12 (g) + O_2 (g) rightarrow CO_2 (g) + H_2 O(g) From the information given on the cover sheet of this exam, calculate the standard heat of reaction, Delta H_reaction^0 for the combustion of one mole of pentane from 6a above. Calculate the mass of water formed in eq. 6a above if 2.73 L of pentane stored at 3.22 atm pressure and 30 degree C is completely burned. How many liters of oxy gen at STP are required to generate 426 kJ of heat. Assume excess pentane.

Explanation / Answer

Hello, from the data provided the only part that can be answered is c)

First, balance the equation:

C5H12 + 8 O2 ---> 5 CO2 + 6 H2O

Then, you need the n of pentane, which can be found using the ideal gas formula PV=nRT

n = PV/RT note: T should be in Kelvin

n = (3.22 atm x 2.73 L) / (0.082 atm.L/K.mol x 303 K)

n = 0.354 mol

and then, by simple cross multiplication:

1 mol C5H12 --------6 mol H2O

0.354 mol C5H12----x

x = (6x0.354)/1 = 2.124 mol of H2O

Finally, for the mass:
1 mol H2O------------18 g H2O

2.124 mol H2O-------x

x = (18 x 2.124)/1 = 38.232 g H2O

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