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The propane fuel (C3H8) used in gas barbeques burns according to a thermochemical equation:

If a pork roast must absorb 1.6 * 103 kJ to fully cook, and if only 10% of the heat produced by the barbeque is actually absorbed by the roast, what mass of CO2 is emitted into the atmosphere during the grilling of the pork roast?

$1.0 \mathrm{~kg}$

04:01

Kevin C.

02:47

Charotte M.

Chemistry 102

Chemistry 101

Chapter 9

Thermochemistry

Thermodynamics

Chemical reactions and Stoichiometry

University of Central Florida

Rice University

Drexel University

University of Maryland - University College

Lectures

03:07

A liquid is a nearly incom…

04:38

A liquid is a state of mat…

01:29

The propane fuel $\left(\m…

01:13

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Many portable gas heaters …

01:53

Consider the combustion of…

03:04

Calculate the enthalpy of …

03:41

Propane $\left(\mathrm{C}_…

03:39

A propane tank on a home b…

03:05

Liquid butane $\left(\math…

03:25

The burner in a gas grill …

All right, so um let's do some barbecue. If we take some propane and oxygen do combustion will create CO. Two and H 20. That will also produce heat. It will release 2044 killing tools of energy. Of course that energy goes into food when we wind up barbecuing. But we all know that not all of the energy goes into the food itself. We know a lot of it gets wasted. We know the barbecue is hot. Um And so not all the heat just goes into the food, but out to the atmosphere as well. So if we know that we need 1.6 times 10 to the three Killah jewels of energy to cook a substance. But we also know that that's 10% efficient on Uh in terms of our barbecue, that means we really need 10 times the amount of energy to get to 100%. So what that really means is we need 1.6 times 10 to the four killing tools of energy to barbecue because Only 10% of it will go into our food and we need 10% of 1.6 times 10 to the fourth or 1.6 times 10 to the third killer jewels. So we're going to use the larger number to figure out how much CO two we will omit. So we're gonna start out with that amount of heat. Yeah. What do we know? Okay, we know from our delta H. That we um wind up using 2044. Kill the jewels of energy. Well we release that for how much co two. If we take a look at the co two, We've got three of them. So we released that amount of energy for three moles of CO two. Now, the other thing before we move on to the last step is this 1.6 times 10 to the fourth. Killing joules of energy. It's going into the food but it has to come out of somewhere. It's coming out of the reaction and out is negative. Yeah. So later these negatives will wind up canceling out Now to get mass of co two. Now we just go to Mulcair mass one mole of CO two. If we add up a carbon into oxygen is 44 0.1 grams. And now if we cancel out our killer jewels, we cancel out our moles of CO two. We're left with grams of CO two. And when we answer this problem, we're going to have to answer to two significant figures. Um Because we only have two significant figures in 1.6 times 10 to the fourth killer jewels. And so when we do the math, we wind up with um 1000 3rd 1033g. But we can't use 1033 because we're only going to two significant figures 1000 g. And if we put a little line over that, that would tell you that second zero is significant. And if you don't know the rule about the line, then what you have to do is have your answer in scientific notation. So you have the two significant figures at 1.0 So 1.8 times 10 to the 3rd g is 1000 g. But to our two significant figures.

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