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3g of a hydrocarbon on combustion in excess of oxygen produces 8.8g of co2 and 5.4g of H2O this data illustrates the law of
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The Law of Conservation of Mass
The law of conservation of mass states that in a chemical reaction, the total mass of the reactants equals the total mass of the products. This means that no matter is created or destroyed during a chemical reaction, only rearranged.
Application of the Law in the Given Data
The combustion of a hydrocarbon in excess oxygen is a chemical reaction that illustrates the law of conservation of mass. In this reaction, the hydrocarbon reacts with oxygen to produce carbon dioxide and water.
Writing the Balanced Chemical Equation
The balanced chemical equation for the combustion of a hydrocarbon in excess oxygen is:
CxHy + (x + y/4)O2 → xCO2 + (y/2)H2O
where x and y are the coefficients of the hydrocarbon and oxygen, respectively.
Calculating the Mass of the Hydrocarbon
From the given data, we know that 8.8g of CO2 and 5.4g of H2O are produced when 3g of the hydrocarbon is combusted. Using the balanced chemical equation, we can calculate the mass of the hydrocarbon as follows:
1 mole of CO2 has a mass of 44g
1 mole of H2O has a mass of 18g
Therefore, the number of moles of CO2 and H2O produced can be calculated as:
moles of CO2 = 8.8g / 44g/mol = 0.2 mol
moles of H2O = 5.4g / 18g/mol = 0.3 mol
From the balanced chemical equation, we know that the number of moles of CO2 produced is equal to the number of moles of the hydrocarbon combusted. Therefore, the number of moles of the hydrocarbon can be calculated as:
moles of hydrocarbon = 0.2 mol
The molar mass of the hydrocarbon can be calculated using its mass and the number of moles:
molar mass of hydrocarbon = 3g / 0.2 mol = 15g/mol
Verification of the Law of Conservation of Mass
Using the balanced chemical equation and the calculated molar mass of the hydrocarbon, we can verify the law of conservation of mass as follows:
Mass of reactants = mass of hydrocarbon + mass of oxygen
Mass of reactants = 3g + (x + y/4)O2
Mass of products = mass of CO2 + mass of H2O
Mass of products = 8.8g + 5.4g
From the balanced chemical equation, we know that the molar ratio of the hydrocarbon to oxygen is 1: (x + y/4). Therefore, the mass of oxygen can be calculated as:
moles of hydrocarbon = 0.2 mol
moles
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3g of a hydrocarbon on combustion in excess of oxygen produces 8.8g of co2 and 5.4g of H2O this data illustrates the law of for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 3g of a hydrocarbon on combustion in excess of oxygen produces 8.8g of co2 and 5.4g of H2O this data illustrates the law of covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3g of a hydrocarbon on combustion in excess of oxygen produces 8.8g of co2 and 5.4g of H2O this data illustrates the law of.
3g of a hydrocarbon on combustion in excess of oxygen produces 8.8g of co2 and 5.4g of H2O this data illustrates the law of for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 3g of a hydrocarbon on combustion in excess of oxygen produces 8.8g of co2 and 5.4g of H2O this data illustrates the law of covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3g of a hydrocarbon on combustion in excess of oxygen produces 8.8g of co2 and 5.4g of H2O this data illustrates the law of.
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