Consider the combustion of propane gas, C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Ho =

Question

Consider the combustion of propane gas, C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Ho = -2,220 kJ/mol Propane (just C3H8) is often used for gas grills. Anyone who has every filled or moved those tanks knows they can get pretty heavy. a) How many grams of propane are in 14 pounds of propane? Use the conversion 1 lb = 454 g. (Express your answers for the next three questions in scientific notation. For example use 2.3e-5 to indicate a number such as 2.3 x 10-5.) grams b) How many moles of propane are in 14 pounds of propane? moles c)How much heat can be obtained by burning 14 pounds of propane? (Remember to look at this from the viewpoint of the surroundings, since the question asks how much heat can be OBTAINED.) kJ What is the enthalpy change for 9 CO2(g) + 12 H2O(l) 3 C3H8(g) + 15 O2(g) (Hint: compare this equation to the combustion equation for propane. kJ

Explanation / Answer

1 lb of propane = 454 g5

So, 14 lb = 6.356e+3

One mole of propane = (12*3 + 1*8) = 36+ 8 = 44 g

So, 14 lb = 6356 g = 6356 / 44 = 144.45 moles of propane

One mole of propane on burning shall give -2,220 kJ

So, 144.45 oles shall give -2,220 * 144.45 = -3.20 MJ of energy shall be released

The enthalpy change for the reverse reaction will be 2,220 kJ per one mole of propane.

Since it's exactly the opposite of tthe combustion reaction.

So, enthalpy change for 9 CO2(g) + 12 H2O(l) 3 C3H8(g) + 15 O2(g) =

= 3 * 2,220 = 6,660 kJ per mole

Hope this helps. Please Rate it ASAP. Thanks.

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