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10ml of gaseous hydrocarbon on combustion give 40ml of CO2 and 50ml of h2o. the hydrocarbon is?
Most Upvoted Answer
10ml of gaseous hydrocarbon on combustion give 40ml of CO2 and 50ml of...
Provided they are all gases then the formula is going to be
C4H10
they key here is that 1 mole of any ideal gass occupies the same volume as 1 mole of any other ideal gas under the same conditions of temperature and pressure. So, since the same volume of any ideal gas has the same number of moles then you can treat the ratio of volumes as the ratio of moles.
the ratio of volumes is
10 : 40 : 50
1 : 4 : 5
all the Carbon from the hydrocarbon is in the CO2
10 ml of hydrocarbon produces 40 ml CO2,
1 : 4 ratio
So 1 mole hydrocarbon has 4 moles Carbon
All the H ends up in the water
10 ml hydrocarbon produces 50 ml water,
1 : 5 ratio,
but also each water has 2 H atoms
So 1 mole hydrocarbon has 10 moles Hydrogen
which gives
C4H10
Community Answer
10ml of gaseous hydrocarbon on combustion give 40ml of CO2 and 50ml of...
Given:
10ml of gaseous hydrocarbon on combustion produce 40ml of CO2 and 50ml of H2O.
To Find:
The hydrocarbon present in the given sample.
Explanation:
Step 1: Determine the moles of CO2 and H2O produced.
We know that 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP). Therefore, we can use the following equations to calculate the moles of CO2 and H2O produced:
Moles of CO2 = Volume of CO2 / Volume of 1 mole of CO2
Moles of H2O = Volume of H2O / Volume of 1 mole of H2O
Given:
Volume of CO2 = 40 ml
Volume of H2O = 50 ml
Calculations:
Moles of CO2 = 40 ml / 22.4 ml/mol = 1.79 moles of CO2
Moles of H2O = 50 ml / 22.4 ml/mol = 2.23 moles of H2O
Step 2: Determine the ratio of moles of CO2 to moles of H2O.
Since the hydrocarbon is composed of carbon and hydrogen, we can use the stoichiometry of combustion to determine the ratio of moles of CO2 to moles of H2O. The balanced equation for the combustion of a hydrocarbon is:
CnHm + (n + m/4) O2 -> n CO2 + m/2 H2O
From the balanced equation, we can see that the ratio of moles of CO2 to moles of H2O is n : m/2.
Step 3: Calculate the ratio of moles of CO2 to moles of H2O.
Given:
Moles of CO2 = 1.79 moles
Moles of H2O = 2.23 moles
Calculations:
Ratio of moles of CO2 to moles of H2O = 1.79 / 2.23 ≈ 0.803
Step 4: Determine the empirical formula of the hydrocarbon.
The empirical formula of a compound represents the simplest whole number ratio of the atoms present in the compound. In this case, the empirical formula of the hydrocarbon can be determined using the ratio of moles of carbon to moles of hydrogen.
Given:
Ratio of moles of CO2 to moles of H2O = 0.803
Calculations:
Assuming we have 1 mole of the hydrocarbon, we can determine the moles of carbon and hydrogen using the ratio from Step 3.
Moles of carbon = 0.803 moles
Moles of hydrogen = 2 * 0.803 moles = 1.606 moles
The empirical formula of the hydrocarbon can be represented as C0.803H1.606.
Step 5: Determine the molecular formula of the hydrocarbon.
To determine the molecular formula, we need the molar mass of the hydrocarbon. We can calculate it by adding the molar masses of carbon and hydrogen.
Given:
Molar mass of carbon = 12.01 g/mol
10ml of gaseous hydrocarbon on combustion produce 40ml of CO2 and 50ml of H2O.
To Find:
The hydrocarbon present in the given sample.
Explanation:
Step 1: Determine the moles of CO2 and H2O produced.
We know that 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP). Therefore, we can use the following equations to calculate the moles of CO2 and H2O produced:
Moles of CO2 = Volume of CO2 / Volume of 1 mole of CO2
Moles of H2O = Volume of H2O / Volume of 1 mole of H2O
Given:
Volume of CO2 = 40 ml
Volume of H2O = 50 ml
Calculations:
Moles of CO2 = 40 ml / 22.4 ml/mol = 1.79 moles of CO2
Moles of H2O = 50 ml / 22.4 ml/mol = 2.23 moles of H2O
Step 2: Determine the ratio of moles of CO2 to moles of H2O.
Since the hydrocarbon is composed of carbon and hydrogen, we can use the stoichiometry of combustion to determine the ratio of moles of CO2 to moles of H2O. The balanced equation for the combustion of a hydrocarbon is:
CnHm + (n + m/4) O2 -> n CO2 + m/2 H2O
From the balanced equation, we can see that the ratio of moles of CO2 to moles of H2O is n : m/2.
Step 3: Calculate the ratio of moles of CO2 to moles of H2O.
Given:
Moles of CO2 = 1.79 moles
Moles of H2O = 2.23 moles
Calculations:
Ratio of moles of CO2 to moles of H2O = 1.79 / 2.23 ≈ 0.803
Step 4: Determine the empirical formula of the hydrocarbon.
The empirical formula of a compound represents the simplest whole number ratio of the atoms present in the compound. In this case, the empirical formula of the hydrocarbon can be determined using the ratio of moles of carbon to moles of hydrogen.
Given:
Ratio of moles of CO2 to moles of H2O = 0.803
Calculations:
Assuming we have 1 mole of the hydrocarbon, we can determine the moles of carbon and hydrogen using the ratio from Step 3.
Moles of carbon = 0.803 moles
Moles of hydrogen = 2 * 0.803 moles = 1.606 moles
The empirical formula of the hydrocarbon can be represented as C0.803H1.606.
Step 5: Determine the molecular formula of the hydrocarbon.
To determine the molecular formula, we need the molar mass of the hydrocarbon. We can calculate it by adding the molar masses of carbon and hydrogen.
Given:
Molar mass of carbon = 12.01 g/mol
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10ml of gaseous hydrocarbon on combustion give 40ml of CO2 and 50ml of h2o. the hydrocarbon is?
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10ml of gaseous hydrocarbon on combustion give 40ml of CO2 and 50ml of h2o. the hydrocarbon is? 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 10ml of gaseous hydrocarbon on combustion give 40ml of CO2 and 50ml of h2o. the hydrocarbon is? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 10ml of gaseous hydrocarbon on combustion give 40ml of CO2 and 50ml of h2o. the hydrocarbon is?.
10ml of gaseous hydrocarbon on combustion give 40ml of CO2 and 50ml of h2o. the hydrocarbon is? 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 10ml of gaseous hydrocarbon on combustion give 40ml of CO2 and 50ml of h2o. the hydrocarbon is? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 10ml of gaseous hydrocarbon on combustion give 40ml of CO2 and 50ml of h2o. the hydrocarbon is?.
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