Problem 1
Question
A quantity of \(10 \mathrm{~g}\) of a hydrocarbon exactly requires \(40 \mathrm{~g}\) oxygen for complete combustion. The products formed are \(\mathrm{CO}_{2}\) and water. When \(\mathrm{CO}_{2}\) gas formed is absorbed completely in lime water, the mass of solution increases by \(27.5 \mathrm{~g}\). What is the mass of water formed in combustion? (a) \(22.5 \mathrm{~g}\) (b) \(27.5 \mathrm{~g}\) (c) \(50 \mathrm{~g}\) (d) \(10 \mathrm{~g}\)
Step-by-Step Solution
Verified Answer
The mass of water formed in combustion is 22.5 g.
1Step 1: Determine the Mass Increase from CO2 Absorption
The mass of the solution after absorbing CO2 increases by 27.5 g. This increase is due to the mass of CO2 being absorbed.
2Step 2: Calculate the Mass of Water Produced
The mass of water produced in the combustion can be determined by subtracting the mass of CO2 absorbed (27.5 g) from the total mass increase when combined with oxygen (40 g).
3Step 3: Apply Law of Conservation of Mass
According to the law of conservation of mass, the total mass before and after the reaction must be equal. Since 10 g of hydrocarbon reacts with 40 g of oxygen, the total mass before combustion is 10 g + 40 g = 50 g. After combustion, this mass is divided into CO2 and water.
4Step 4: Compute the Mass of Water
Subtract the mass of CO2 absorbed by the lime water (27.5 g) from the total mass after combustion (50 g) to find the mass of water formed.
Key Concepts
Mass Conservation LawChemical Reactions in CombustionStoichiometry Calculations
Mass Conservation Law
Imagine you're baking cookies. You weigh out your ingredients, mix them together, and then bake your cookies. After baking, your cookies still weigh the same as the sum of all your ingredients, right? This is what the mass conservation law is all about, but in chemical reactions. It states that the total mass of reactants in a chemical reaction must equal the total mass of the products.
This law is crucial when we're trying to figure out what happens during a reaction, like the combustion of a hydrocarbon. In the given exercise, a hydrocarbon weighing 10 g reacts with 40 g of oxygen. According to the mass conservation law, the combined mass of carbon dioxide (CO2) and water produced must equal that 50 g. This concept is the foundation upon which the whole problem is solved, ensuring that we account for every gram of reactant and product.
This law is crucial when we're trying to figure out what happens during a reaction, like the combustion of a hydrocarbon. In the given exercise, a hydrocarbon weighing 10 g reacts with 40 g of oxygen. According to the mass conservation law, the combined mass of carbon dioxide (CO2) and water produced must equal that 50 g. This concept is the foundation upon which the whole problem is solved, ensuring that we account for every gram of reactant and product.
Chemical Reactions in Combustion
Combustion is like a dance between molecules where oxygen is the dance partner everyone wants. Specifically, it's a chemical reaction where a substance combines with oxygen to release energy and produce new substances. For hydrocarbons—a category that includes a lot of different molecules made up of hydrogen and carbon—that dance leads to the production of carbon dioxide and water.
In our exercise scenario, the hydrocarbon and oxygen are the 'dancers', and the CO2 and water are the 'dance outcomes'. As in any proper chemical dance, nothing gets lost; everything just changes partners. The mass that goes into the dance (the reactants) is the same mass that comes out of it (the products), which is why knowing that the CO2 absorbed by lime water increases its mass by 27.5 g helps us figure out what else was made in the reaction.
In our exercise scenario, the hydrocarbon and oxygen are the 'dancers', and the CO2 and water are the 'dance outcomes'. As in any proper chemical dance, nothing gets lost; everything just changes partners. The mass that goes into the dance (the reactants) is the same mass that comes out of it (the products), which is why knowing that the CO2 absorbed by lime water increases its mass by 27.5 g helps us figure out what else was made in the reaction.
Stoichiometry Calculations
Stoichiometry is like the recipe for a chemical reaction. It tells us how much of each ingredient we need and how much product we'll end up with. It's all about the quantities and proportions of reactants and products in a chemical reaction. Stoichiometry calculations are the steps we follow to ‘cook up’ our chemical recipes just right.
In the exercise solution, we use stoichiometry to work out the mass of water produced. We know the initial 'ingredients' (the hydrocarbon and oxygen) and one of the 'dishes' made (the CO2). To find the mass of the other product (water), we do some simple subtraction, guided by stoichiometry principles. We take the total mass after the reaction (50 g from both hydrocarbon and oxygen) and subtract the mass of CO2 to find the mass of the water, ensuring we have accounted for all the products and obeyed the law of mass conservation.
In the exercise solution, we use stoichiometry to work out the mass of water produced. We know the initial 'ingredients' (the hydrocarbon and oxygen) and one of the 'dishes' made (the CO2). To find the mass of the other product (water), we do some simple subtraction, guided by stoichiometry principles. We take the total mass after the reaction (50 g from both hydrocarbon and oxygen) and subtract the mass of CO2 to find the mass of the water, ensuring we have accounted for all the products and obeyed the law of mass conservation.
Other exercises in this chapter
Problem 1
A sample of clay contains \(50 \%\) silica and \(10 \%\) water. The sample is partially dried by which it loses \(8 \mathrm{~g}\) water. If the percentage of si
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When a mixture of aluminium powder and iron (III) oxide is ignited, it produces molten iron and aluminium oxide. In an experiment, \(5.4 \mathrm{~g}\) of alumin
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Assume that sodium atoms are spheres of radius \(0.2 \mathrm{~nm}\) and that they are lined up side by side. How many miles, in length, is the line of atoms pre
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