The propane fuel (C 3 H 8 ) used in gas barbeques burns according to the thermochemical equation: C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(g) ΔH° rxn = –2044 kJ If a pork roast must absorb 1.6×10 3 kJ to fully cook, and if only 10% of the heat produced by the barbeque is actually absorbed by the roast, what mass of CO 2 is emitted into the atmosphere during the grilling of the pork roast?

All right. Hi, everyone. So this question says that the propane fuel C three H eight used in gas barbecues burns according to the thermo chemical equation where one mole of propane reacts with five moles of 02 to produce three moles of carbon dioxide and four moles of water. Delta H Standard of the reaction is equal to negative 2044 keto joules. If a pork roast must absorb 1.6 multiplied by 10th of the third kilo jewels to be fully cooked. And if only 10% of the heat produced by the barbecue is absorbed by the roast, what mass of carbon dioxide is emitted into the atmosphere during the grilling of the pork roast. So in this case, in order to understand the solution for this question, we have to recall the fact that enthalpy can be utilized in calculations like this one using stoichiometry principles because we're focusing on carbon dioxide. For this question, we can say the three moles of carbon dioxide as seen in the balanced chemical equation. Correlates to the value for Delta H standard which is 2044 kilojoules. We can treat this as a conversion factor. And because we're to find the mass of carbon dioxide, we also have to recall the molar mass of carbon dioxide, which is equal to 44.01 g per mole. So we have almost all of the information we need, but there's one more bit of information that we have to analyze a little bit further. That is the efficiency of the barbecue itself. Only 10% of the heat produced is absorbed by the roast. So what we can say is that for every 100 kg jewels produce only 10 kg joules is absorbed. So notice what the question is starting off with the pork roast must absorb 1.6 multiplied by 10th, a third kilo jewels. So for my given, which is the first piece of information I would write down, I would write 1.6 multiplied by 10 to the third keto Joel absorbed. So I can relate this then to the amount of energy that has to be put into the barbecue, which is going to be more than the amount that's going to be absorbed due to the efficiency. Now, in my conversion factor, recall that the idea is to make sure that your starting units cancel out. So because of this, when using the efficiency of the barbecue as a conversion factor, the kilo jewels absorbed goes in the denominator, which means that kilo jewel produced must be the numerator. So that's 100 kilojoules produced in the numerator and 10 kilojoules absorbed in the denominator, this means that kilojoules absorb absorbed, excuse me, cancel out. So now that we're taking into consideration how much energy has to be generated overall or produced overall, we can relate this back to the delta H standard of the reaction. So I can introduce now, my second conversion factor which relates delta H standard to the moles of carbon dioxide. So because kilo jewels are my current units. In my second conversion factor, the kilo jewels associated to carbon dioxide should be in the denominator. So that's three moles of co two divided by 2044 kg jewels produced that tequila jules produced. So once again, units of kilo jewels cancel out entirely. And now our answer is going to be expressed as moles of carbon dioxide. So because our target units are grams of carbon dioxide, our final conversion factor is going to be with the molar mass of carbon dioxide. So my last conversion factor here would place moles of carbon dioxide in the denominator because the idea is to ensure that those units cancel out. So that's 44.01 g of CO2 per one mole of CO2. So now to go ahead and evaluate this expression, starting off with the quantity given you are to go ahead and multiply all numerators and divide all denominators. Once that's done, this equals 1033.5 grams of carbon dioxide, which after rounding to two significant figures equals 1.0 multiplied by 10 to the third power grams of carbon dioxide. And there you have it. That is our final answer. And so with that being said, thank you so very much for watching and I hope you found this helpful.

Ch.6 - Thermochemistry

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Tro 4th Edition

Ch.6 - Thermochemistry

Problem 63

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Chapter 6, Problem 63

The propane fuel (C₃H₈) used in gas barbeques burns according to the thermochemical equation: C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g) ΔH°_rxn = –2044 kJ 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?

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Calculate the total amount of heat required to cook the pork roast considering only 10% efficiency. This is done by dividing the heat needed by the roast (1.6 * 10^3 kJ) by 0.10.

Determine the amount of propane (C3H8) needed to produce the calculated total heat. Use the given ΔH°rxn (-2044 kJ per mole of C3H8) and calculate the moles of C3H8 required by dividing the total heat by -2044 kJ/mol.

Using the stoichiometry of the balanced chemical equation, calculate the moles of CO2 produced. For every mole of C3H8 burned, 3 moles of CO2 are produced.

Convert the moles of CO2 to mass. Use the molar mass of CO2 (approximately 44.01 g/mol) and multiply it by the number of moles of CO2 calculated in the previous step.

The result from the previous step gives the mass of CO2 emitted into the atmosphere during the grilling of the pork roast.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Thermochemical Equations

Thermochemical equations represent the heat changes associated with chemical reactions. They include the enthalpy change (ΔH) which indicates whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). In this case, the combustion of propane releases -2044 kJ of energy, which is crucial for calculating how much energy is available for cooking the pork roast.

Heat Transfer Efficiency

Heat transfer efficiency refers to the percentage of heat produced that is effectively utilized for a specific purpose. In this scenario, only 10% of the heat generated by the barbeque is absorbed by the pork roast. This efficiency is vital for determining the actual amount of energy available for cooking, which influences the calculations for the mass of CO2 produced.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced equations. It allows us to relate the amount of propane burned to the amount of carbon dioxide produced. By using the stoichiometric coefficients from the combustion reaction, we can determine how much CO2 is emitted based on the energy absorbed by the roast.