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) \(10 \mathrm{~g}\)

Short answer

Answer: The mass of water formed in the complete combustion of the hydrocarbon is \(22.5 \mathrm{~g}\).

Step-by-step solution

Write the balanced chemical equation

For any hydrocarbon (represented by general formula CxHy), the complete combustion reaction can be represented as: CxHy + O2 -> n*CO2 + m*H2O Here, n and m represent the stoichiometric coefficients that we don't yet know. Balancing the equation is not important in this particular problem since it doesn't involve finding the unknown as part of the exercise.

Determine the mass of CO2 absorbed in lime water

We know that the mass of the lime water solution increases by 27.5 g when CO2 is absorbed. This means the mass of CO2 formed in the reaction is also 27.5 g.

Calculate the total mass of products formed during combustion

According to the Law of Conservation of Mass, the mass of all reactants together must be equal to the mass of all products together. So, in this case, Mass of reactants = Mass of hydrocarbon + Mass of oxygen = 10 g + 40 g = 50 g Therefore, the total mass of products (CO2 and H2O) formed during the combustion = 50 g

Determine the mass of water formed in the reaction

We now know the mass of CO2 and the total mass of products. We can find the mass of water formed by subtracting the mass of CO2 from the total mass of products. Mass of water formed = Mass of products - Mass of CO2 = 50 g - 27.5 g = 22.5 g So the mass of water formed in the combustion is 22.5 g. Hence, the correct answer is (a) \(22.5 \mathrm{~g}\).

Key concepts

Stoichiometry

Stoichiometry is the art of balancing elements in chemical reactions to understand the proportions of reactants and products. It helps us figure out exactly how much of each chemical substance is involved in a reaction by using the mole concept. In this case, a hydrocarbon reacts with oxygen to produce carbon dioxide and water.
To determine the stoichiometry of the equation, you’ll use the coefficients in these reactions, which are based on balanced chemical equations. Here, the hydrocarbon's stoichiometric coefficients indicate how many molecules of oxygen will react with it to produce specific amounts of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\).
This idea assumes you can scale reactions up or down based on these numbers. Even though the balancing of the equation may not directly be involved in this problem, it is crucial for understanding the chemistry behind how much carbon dioxide and water are produced.

• The coefficients give proportions between reactants and products.
• Ensures that amount equations accurately reflect the actual reaction.
• Determines usage of molar masses to calculate real masses used/discovered in a laboratory.

Law of Conservation of Mass

The Law of Conservation of Mass states that in a closed system, mass is conserved; it cannot be created or destroyed. It is a fundamental principle in chemistry.
In the combustion reaction given, the total mass of the reactants (hydrocarbon and oxygen) is equal to the total mass of the products (carbon dioxide and water). Calculating this helps us determine the unknown mass of water produced.
For this problem, adding together the masses of the hydrocarbon and oxygen gives us the total mass of the reactants. According to the conservation principle, this should be equal to the mass of carbon dioxide and water formed:

• Reactant mass is calculated by adding the hydrocarbon (10 g) and oxygen (40 g).
• The product mass is split between carbon dioxide and water, matching the reactant mass at 50 g.
• Helps verify calculations are properly balanced and align with real-world chemical reactions.

This law is an essential check on our work, proving that calculations must align with physical reality.

Chemical Equations

Chemical equations are a shorthand notation that provide vital information about chemical reactions. They include:
• The substances involved (reactants and products).
• The physical state of each reactant and product.
• Coefficients showing the stoichiometric ratios of how substances react.
In a balanced chemical equation for the combustion reaction of a hydrocarbon, carbon dioxide \((\mathrm{CO}_2)\) and water \((\mathrm{H}_2\mathrm{O})\) result from combining with oxygen \((\mathrm{O}_2)\).
For this exercise, the equation reflects:

• The reactants: Hydrocarbon and oxygen involved in burning.
• The products: The outputs are carbon dioxide, absorbed by lime water, and water.

The chemical equation gives us a framework for understanding stoichiometry and the conservation of mass, illuminating how substance masses relate to each other in real-life reactions.
While we may not need to balance every equation in detail here, recognizing their components helps us predict what will form and why, as seen by the exact calculation of water left after carbon dioxide absorption.