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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.?
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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
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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.?
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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.?.
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