# Question #c91cf

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

Your question is missing a very important piece of information - the *enthalpy change of combustion* for **liquified petroleum gas**, or **LPG**.

WIthout knowing how much heat is released when LPG undergoes combustion you cannot determine how much LPG would be needed to heat the water.

I'm assuming that the question given to you included a chemical equation. To keep things simple, you can represent LPG as either propane,

I'll show you how to solve the problem for *propane*. The combustion reaction of propane is

#"C"_3"H"_text(8(g]) + 5"O"_text(2(g]) ->3"CO"_text(2(g]) + 4"H"_2"O"_text((l])#

The standard enthalpy change of combustion for propane is

Here's you strategy for this problem. Use the given volume of water to determine how much heat would be needed to increase the water's temperature from

The equation that establishes a relationship between heat and temperature increase looks like this

#color(blue)(q = m * c * DeltaT)" "# , where

*mass* of the water, equal to

SInce no information about water's density was given, you can assume it to be equal to

This means that you have

#1.5color(red)(cancel(color(black)("L"))) * (1color(red)(cancel(color(black)("kg"))))/(1color(red)(cancel(color(black)("L")))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 1.5 * 10^3"g"#

of water in your sample.

The amount of heat needed to increase the temperature of the water will be

#q = 1.5 * 10^3color(red)(cancel(color(black)("g"))) * 4.184"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (100 - 25)color(red)(cancel(color(black)(""^@"C")))#

#q = 470.7 * 10^3"J" = "470.7 kJ"#

Now look at the enthalpy of combustion for that reaction. If **one mole** of propane gives off

#470.7color(red)(cancel(color(black)("kJ"))) * ("1 mole C"_3"H"_8)/(2202color(red)(cancel(color(black)("kJ")))) = "0.2138 moles C"_3"H"_8#

to produce the amount of heat you need to heat the water.

**However**, keep in mind that this amount would be needed if

You know that that's not the case, since only

This means that you would need *more propane* to make sure that you get this much heat

#0.2138 * 100/16 = "1.336 moles C"_3"H"_8#

Finally, to get the mass of propane needed, use its molar mass

#1.336color(red)(cancel(color(black)("moles"))) * "44.1 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("58.9 g C"_3"H"_8)#