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LP gas burns according to the exothermic reaction:

What mass of LP gas is necessary to heat 1.5 L of water from room temperature (25.0 C) to boiling (100.0 C)? Assume that during heating, 15% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings.

67.3

Chemistry 102

Chemistry 101

Chapter 9

Thermochemistry

Thermodynamics

Chemical reactions and Stoichiometry

Drexel University

Brown University

Lectures

00:42

In thermodynamics, the zer…

01:47

A spontaneous process is o…

01:43

LP gas burns according to …

06:44

01:12

Carbon dioxide sublimes (c…

02:23

When 3.00 L of propane gas…

04:23

The combustion of one mole…

01:58

If all the energy obtained…

00:48

A cook wants to heat $1.35…

04:59

How much heat is required …

02:07

When 2.50 g of methane bur…

0:00

Calculate the amount of he…

01:35

The enthalpy of sublimatio…

02:41

How much heat evolves when…

04:52

02:50

01:16

When 5.00 g of a compound …

02:05

The hot-water heater descr…

01:13

02:46

Dissolving 3.0 $\mathrm{g}…

05:39

An ideal gas is heated fro…

00:50

The heat of combustion of …

So in this problem, we're given the chemical reaction for the combustion of propane. See three h eight gas plus five o to gas forms. Three co two gas plus for H two o gas. And this has a delta h not of reaction of negative twenty forty four. Kill Jules, and we want to find what massive propane is needed to heat one point five litres of water from twenty five degrees Celsius to boiling at a hundred degrees Celsius. And so we can do this with the equation. Q Equal cm Delta T. Where C is the specific heat. Em is mass and Delta teams the change in temperature. And so the specific heat of water is four point one eight four. The mass. We have one point five leaders, which, since the density ofwater is one gram per mil leader that gives us fifteen hundred grams. So we multiply by fifteen hundred grams and then we multiplied by the change in temperature, final minus initial. And that gives us four hundred seventy one. Kill it, Jules. And so I multiply these out and then converted to kill a Jules. And so now that we have that, we can find the number of moles we need of propane and then thus find the number of grams. And so, since we're only using fifteen percent of the heat generated by this reaction, and the rest is being lost to the atmosphere, our fellow are effective. Delta Age is our delta h times that fifteen percent. And that gives us negative three hundred seven killer Jules Permal, because is the standard conditions. Assume that this is the heat Permal. And so now that we have that we can take our four hundred seventy one killer jewels of heat that we need and divide that by r three hundred seven killer Jules Permal that we're getting and find out that we need to burn one point five three moles off propane. So now we can just multiply by the mother mass of propane in order to get that into grams. And we find out that we have sixty three point seven grams of propane and that is our final answer

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