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mass of ammonia produced on mixing 56 gram of Nitrogen with 10 gram of hydrogen is a) 34.5 g b) 56.7 g c) 68 g d) 63.7 g
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mass of ammonia produced on mixing 56 gram of Nitrogen with 10 gram of...
Calculation of Mass of Ammonia Produced on Mixing 56 g Nitrogen and 10 g Hydrogen
Step 1: Write the balanced chemical equation for the reaction between nitrogen and hydrogen to form ammonia.
N2 + 3H2 → 2NH3
Step 2: Calculate the number of moles of nitrogen (N2) and hydrogen (H2) used in the reaction.
Number of moles of N2 = 56 g / 28 g/mol = 2 mol
Number of moles of H2 = 10 g / 2 g/mol = 5 mol
Step 3: Determine the limiting reactant in the reaction.
The limiting reactant is the reactant that is completely consumed in the reaction and determines the amount of product formed. To determine the limiting reactant, we compare the number of moles of each reactant with the stoichiometric coefficients in the balanced chemical equation.
From the balanced chemical equation, we see that 1 mol of N2 reacts with 3 mol of H2 to form 2 mol of NH3. Therefore, the amount of NH3 that can be produced from 2 mol of N2 is:
2 mol N2 × (2 mol NH3 / 1 mol N2) = 4 mol NH3
Similarly, the amount of NH3 that can be produced from 5 mol of H2 is:
5 mol H2 × (2 mol NH3 / 3 mol H2) = 3.33 mol NH3
Since the amount of NH3 that can be produced from N2 is greater than the amount that can be produced from H2, H2 is the limiting reactant.
Step 4: Calculate the number of moles of NH3 produced.
From the balanced chemical equation, we see that 3 mol of H2 reacts with 1 mol of NH3. Therefore, the amount of NH3 that can be produced from 5 mol of H2 is:
5 mol H2 × (1 mol NH3 / 3 mol H2) = 1.67 mol NH3
Step 5: Calculate the mass of NH3 produced.
Mass of NH3 = number of moles of NH3 × molar mass of NH3
Mass of NH3 = 1.67 mol × 17 g/mol = 28.4 g
Therefore, the mass of ammonia produced on mixing 56 g of nitrogen with 10 g of hydrogen is 28.4 g.
Answer: (a) 34.5 g
Step 1: Write the balanced chemical equation for the reaction between nitrogen and hydrogen to form ammonia.
N2 + 3H2 → 2NH3
Step 2: Calculate the number of moles of nitrogen (N2) and hydrogen (H2) used in the reaction.
Number of moles of N2 = 56 g / 28 g/mol = 2 mol
Number of moles of H2 = 10 g / 2 g/mol = 5 mol
Step 3: Determine the limiting reactant in the reaction.
The limiting reactant is the reactant that is completely consumed in the reaction and determines the amount of product formed. To determine the limiting reactant, we compare the number of moles of each reactant with the stoichiometric coefficients in the balanced chemical equation.
From the balanced chemical equation, we see that 1 mol of N2 reacts with 3 mol of H2 to form 2 mol of NH3. Therefore, the amount of NH3 that can be produced from 2 mol of N2 is:
2 mol N2 × (2 mol NH3 / 1 mol N2) = 4 mol NH3
Similarly, the amount of NH3 that can be produced from 5 mol of H2 is:
5 mol H2 × (2 mol NH3 / 3 mol H2) = 3.33 mol NH3
Since the amount of NH3 that can be produced from N2 is greater than the amount that can be produced from H2, H2 is the limiting reactant.
Step 4: Calculate the number of moles of NH3 produced.
From the balanced chemical equation, we see that 3 mol of H2 reacts with 1 mol of NH3. Therefore, the amount of NH3 that can be produced from 5 mol of H2 is:
5 mol H2 × (1 mol NH3 / 3 mol H2) = 1.67 mol NH3
Step 5: Calculate the mass of NH3 produced.
Mass of NH3 = number of moles of NH3 × molar mass of NH3
Mass of NH3 = 1.67 mol × 17 g/mol = 28.4 g
Therefore, the mass of ammonia produced on mixing 56 g of nitrogen with 10 g of hydrogen is 28.4 g.
Answer: (a) 34.5 g
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mass of ammonia produced on mixing 56 gram of Nitrogen with 10 gram of hydrogen is a) 34.5 g b) 56.7 g c) 68 g d) 63.7 g
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mass of ammonia produced on mixing 56 gram of Nitrogen with 10 gram of hydrogen is a) 34.5 g b) 56.7 g c) 68 g d) 63.7 g for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about mass of ammonia produced on mixing 56 gram of Nitrogen with 10 gram of hydrogen is a) 34.5 g b) 56.7 g c) 68 g d) 63.7 g covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for mass of ammonia produced on mixing 56 gram of Nitrogen with 10 gram of hydrogen is a) 34.5 g b) 56.7 g c) 68 g d) 63.7 g.
mass of ammonia produced on mixing 56 gram of Nitrogen with 10 gram of hydrogen is a) 34.5 g b) 56.7 g c) 68 g d) 63.7 g for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about mass of ammonia produced on mixing 56 gram of Nitrogen with 10 gram of hydrogen is a) 34.5 g b) 56.7 g c) 68 g d) 63.7 g covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for mass of ammonia produced on mixing 56 gram of Nitrogen with 10 gram of hydrogen is a) 34.5 g b) 56.7 g c) 68 g d) 63.7 g.
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