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3.1 *10²³ molecule of N2 react with 8gram of H2 . Find out the maximum amount of ammonia formed.?
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3.1 *10²³ molecule of N2 react with 8gram of H2 . Find out the maximum...
Given data:
- Number of molecules of N2 = 3.1 * 10^23
- Mass of H2 = 8g
Step 1: Determining the limiting reactant
- Calculate the moles of N2:
Moles = Number of molecules / Avogadro's number
Moles = 3.1 * 10^23 / 6.022 * 10^23
Moles = 0.515 moles
- Calculate the moles of H2:
Moles = Mass / Molar mass
Molar mass of H2 = 2g/mol
Moles = 8g / 2g/mol
Moles = 4 moles
- Since N2 has fewer moles than H2, N2 is the limiting reactant.
Step 2: Finding the maximum amount of NH3 formed
- The balanced chemical equation is:
N2 + 3H2 -> 2NH3
- From the equation, 1 mole of N2 produces 2 moles of NH3
- Moles of NH3 produced = Moles of N2 * 2
Moles = 0.515 moles * 2
Moles = 1.03 moles
- Calculate the mass of NH3 produced:
Mass = Moles * Molar mass
Molar mass of NH3 = 17g/mol
Mass = 1.03 moles * 17g/mol
Mass = 17.51g
Conclusion:
- The maximum amount of NH3 formed when 3.1 * 10^23 molecules of N2 react with 8g of H2 is 17.51g. N2 is the limiting reactant, and the amount of NH3 formed is based on the amount of N2 present.
- Number of molecules of N2 = 3.1 * 10^23
- Mass of H2 = 8g
Step 1: Determining the limiting reactant
- Calculate the moles of N2:
Moles = Number of molecules / Avogadro's number
Moles = 3.1 * 10^23 / 6.022 * 10^23
Moles = 0.515 moles
- Calculate the moles of H2:
Moles = Mass / Molar mass
Molar mass of H2 = 2g/mol
Moles = 8g / 2g/mol
Moles = 4 moles
- Since N2 has fewer moles than H2, N2 is the limiting reactant.
Step 2: Finding the maximum amount of NH3 formed
- The balanced chemical equation is:
N2 + 3H2 -> 2NH3
- From the equation, 1 mole of N2 produces 2 moles of NH3
- Moles of NH3 produced = Moles of N2 * 2
Moles = 0.515 moles * 2
Moles = 1.03 moles
- Calculate the mass of NH3 produced:
Mass = Moles * Molar mass
Molar mass of NH3 = 17g/mol
Mass = 1.03 moles * 17g/mol
Mass = 17.51g
Conclusion:
- The maximum amount of NH3 formed when 3.1 * 10^23 molecules of N2 react with 8g of H2 is 17.51g. N2 is the limiting reactant, and the amount of NH3 formed is based on the amount of N2 present.
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3.1 *10²³ molecule of N2 react with 8gram of H2 . Find out the maximum amount of ammonia formed.?
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3.1 *10²³ molecule of N2 react with 8gram of H2 . Find out the maximum amount of ammonia formed.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 3.1 *10²³ molecule of N2 react with 8gram of H2 . Find out the maximum amount of ammonia formed.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3.1 *10²³ molecule of N2 react with 8gram of H2 . Find out the maximum amount of ammonia formed.?.
3.1 *10²³ molecule of N2 react with 8gram of H2 . Find out the maximum amount of ammonia formed.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 3.1 *10²³ molecule of N2 react with 8gram of H2 . Find out the maximum amount of ammonia formed.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3.1 *10²³ molecule of N2 react with 8gram of H2 . Find out the maximum amount of ammonia formed.?.
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