JEE Exam > JEE Questions > Calculate the number of millilitres of NH3 so...
Start Learning for Free
Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.?
Most Upvoted Answer
Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)conta...
Calculation of NH3 solution required for precipitation of Fe(OH)3
Given:
- Density of NH3 solution (d) = 0.986 g/mL
- NH3 concentration = 2.5% by mass
- Mass of sample = 0.8 g
- Fe2O3 concentration in sample = 50%
Step 1: Calculation of mass of Fe2O3 in sample
- Mass of Fe2O3 = 0.8 g x (50/100) = 0.4 g
Step 2: Calculation of moles of Fe2O3 in sample
- Molar mass of Fe2O3 = 2 x 55.85 + 3 x 16.00 = 159.70 g/mol
- Moles of Fe2O3 = mass/molar mass = 0.4 g/159.70 g/mol = 0.00251 mol
Step 3: Calculation of moles of Fe(OH)3 formed
- Balanced equation: Fe2O3 + 3NH3 + 3H2O → 2Fe(OH)3 + 3NH4+
- From the equation, 1 mole of Fe2O3 produces 2 moles of Fe(OH)3
- Moles of Fe(OH)3 formed = 0.00251 mol x 2 = 0.00502 mol
Step 4: Calculation of moles of NH3 required
- From the balanced equation, 3 moles of NH3 react with 1 mole of Fe2O3 to form 2 moles of Fe(OH)3
- Moles of NH3 required = 3/1 x 0.00251 mol = 0.00753 mol
Step 5: Calculation of volume of NH3 solution required
- Mass of NH3 in solution = 0.986 g/mL x 0.025 g/g x volume (mL)
- Mass of NH3 required = moles x molar mass = 0.00753 mol x 17.03 g/mol = 0.128 g
- Volume of NH3 solution required = mass of NH3 required/density of NH3 solution = 0.128 g/0.986 g/mL = 0.130 mL
Therefore, 0.130 mL of NH3 solution with 2.5% by mass NH3 is required to precipitate Fe(OH)3 from a 0.8 g sample containing 50% Fe2O3.
Given:
- Density of NH3 solution (d) = 0.986 g/mL
- NH3 concentration = 2.5% by mass
- Mass of sample = 0.8 g
- Fe2O3 concentration in sample = 50%
Step 1: Calculation of mass of Fe2O3 in sample
- Mass of Fe2O3 = 0.8 g x (50/100) = 0.4 g
Step 2: Calculation of moles of Fe2O3 in sample
- Molar mass of Fe2O3 = 2 x 55.85 + 3 x 16.00 = 159.70 g/mol
- Moles of Fe2O3 = mass/molar mass = 0.4 g/159.70 g/mol = 0.00251 mol
Step 3: Calculation of moles of Fe(OH)3 formed
- Balanced equation: Fe2O3 + 3NH3 + 3H2O → 2Fe(OH)3 + 3NH4+
- From the equation, 1 mole of Fe2O3 produces 2 moles of Fe(OH)3
- Moles of Fe(OH)3 formed = 0.00251 mol x 2 = 0.00502 mol
Step 4: Calculation of moles of NH3 required
- From the balanced equation, 3 moles of NH3 react with 1 mole of Fe2O3 to form 2 moles of Fe(OH)3
- Moles of NH3 required = 3/1 x 0.00251 mol = 0.00753 mol
Step 5: Calculation of volume of NH3 solution required
- Mass of NH3 in solution = 0.986 g/mL x 0.025 g/g x volume (mL)
- Mass of NH3 required = moles x molar mass = 0.00753 mol x 17.03 g/mol = 0.128 g
- Volume of NH3 solution required = mass of NH3 required/density of NH3 solution = 0.128 g/0.986 g/mL = 0.130 mL
Therefore, 0.130 mL of NH3 solution with 2.5% by mass NH3 is required to precipitate Fe(OH)3 from a 0.8 g sample containing 50% Fe2O3.
Community Answer
Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)conta...
10.34ML
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.?
Question Description
Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.?.
Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.?.
Solutions for Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.? defined & explained in the simplest way possible. Besides giving the explanation of
Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.?, a detailed solution for Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.? has been provided alongside types of Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.? theory, EduRev gives you an
ample number of questions to practice Calculate the number of millilitres of NH3 solution(d=0. 986g/ml)contain 2.5% by mass NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8g sample that contains 50% Fe2O3.? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup