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A new grill has a mass of \(30.0 \mathrm{~kg}\). You put \(3.0 \mathrm{~kg}\) of charcoal in the grill. You burn all the charcoal and the grill has a mass of \(30.0 \mathrm{~kg}\). What is the mass of the gases given off? Explain.

Short Answer

Expert verified
The mass of the gases given off during the burning process is 3.0 kg, as determined by applying the law of conservation of mass and calculating the difference between the total mass before (33.0 kg) and after (30.0 kg) the burning process.

Step by step solution

01

Identify given information

The initial mass of the grill is 30.0 kg, the mass of charcoal before burning is 3.0 kg, and the mass of the grill after burning all the charcoal is also 30.0 kg.
02

Apply the law of conservation of mass

According to the law of conservation of mass, the total mass before the burning process should be equal to the total mass after the burning process. So, let's find the total mass before burning: Total mass before = Mass of grill + Mass of charcoal Total mass before = 30.0 kg + 3.0 kg = 33.0 kg
03

Calculate the mass of gases given off

The total mass after burning the charcoal is the mass of the grill (since it's stated that all charcoal has been burned). This means that the mass after burning includes the mass of grill and the mass of gases given off: Total mass after = Mass of grill + Mass of gases given off We know the mass of the grill after burning is 30.0 kg. So, we can rearrange the equation to find the mass of gases given off: Mass of gases given off = Total mass after - Mass of grill Mass of gases given off = 33.0 kg - 30.0 kg = 3.0 kg Thus, the mass of the gases given off during the burning process is 3.0 kg.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Chemical Reactions
In the context of the exercise, chemical reactions are at the core of what's happening when charcoal is burned in a grill. A chemical reaction is a process where reactants transform into products through the breaking and forming of chemical bonds. In the case of burning charcoal, carbon in the charcoal reacts with oxygen in the air to form carbon dioxide. This process releases energy, which we perceive as heat suitable for grilling.

During chemical reactions, substances can change states, like from solid charcoal to gaseous carbon dioxide, but the atoms involved remain the same. The type and number of atoms before and after the reaction stay consistent; they are just rearranged into new molecules. This idea is fundamental, because it links directly to why the mass of the substances involved in a chemical reaction doesn't simply disappear—they are conserved, which leads us to the law of mass conservation.
The Law of Mass Conservation Explained
The law of conservation of mass states that in a closed system subject to no external forces, the mass of the system remains constant over time. It's a principle credited to Antoine Lavoisier, the father of modern chemistry, and it implies that mass can neither be created nor destroyed.

In the problem, even though the charcoal turns into gas and seems to 'disappear,' according to the law of mass conservation, the mass of the gases released must equal the mass of the original charcoal. With this principle, you can confidently conclude that if the grill’s mass remains at 30.0 kg after burning 3.0 kg of charcoal, then the mass of the gas produced must also be 3.0 kg. The mass conservation law is a crucial foundation in understanding chemical reactions quantitatively and is closely related to the concept of stoichiometry.
Stoichiometry and Its Role in Chemical Equations
Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. It is a key concept which ensures that the law of conservation of mass is upheld in chemical equations.

When using stoichiometry, chemists employ a 'mole' concept to balance chemical equations. One mole represents Avogadro's number of particles, such as atoms or molecules. By using the relative atomic or molecular masses, scientists can predict the masses of products formed from a known amount of reactants, or vice versa.

In our exercise, if we knew the chemical formula of the gases produced by burning charcoal, we could use stoichiometry to predict the exact mass of the gas evolved. Without diving into the actual equations and molar masses of the products, the solution already demonstrates a simple stoichiometric principle: equal mass of reactants will produce an equal mass of products.

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Most popular questions from this chapter

Boron consists of two isotopes, \({ }^{10} \mathrm{~B}\) and \({ }^{11} \mathrm{~B}\). Chlorine also has two isotopes, \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\). Consider the mass spectrum of \(\mathrm{BCl}_{3}\). How many peaks would be present, and what approximate mass would each peak correspond to in the \(\mathrm{BCl}_{3}\) mass spectrum?

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. A 752 g sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{~g}\) of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(\mathrm{g}) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of \(0.157 \mathrm{~g}\) of the compound produced \(0.213 \mathrm{~g}\) \(\mathrm{CO}\), and \(0.0310 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that \(0.103 \mathrm{~g}\) of the compound produces \(0.0230 \mathrm{~g} \mathrm{NH}_{3} .\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Calculate the percent composition by mass of the following compounds that are important starting materials for synthetic polymers: a. \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2}\) (acrylic acid, from which acrylic plastics are made) b. \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{2}\) (methyl acrylate, from which Plexiglas is made) c. \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}\) (acrylonitrile, from which Orlon is made)

An element \(\mathrm{X}\) forms both a dichloride \(\left(\mathrm{XCl}_{2}\right)\) and a tetrachloride \(\left(\mathrm{XCl}_{4}\right) .\) Treatment of \(10.00 \mathrm{~g} \mathrm{XCl}_{2}\) with excess chlorine forms \(12.55 \mathrm{~g} \mathrm{XCl}_{4}\). Calculate the atomic mass of \(\mathrm{X}\), and identify \(\underline{X}\)

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