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1000 oxygen at NTP were passed through Siemens ozoniser when the resulting volume 888 ml at NTP. This quantity ozonized oxygen is passed through excess of potassium iodide solution.what is the weight of iodine liberated?? Ans is 2.54gm anyone explane?
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1000 oxygen at NTP were passed through Siemens ozoniser when the resul...
Given Information:
- Volume of oxygen at NTP = 1000 mL
- Volume of resulting ozonized oxygen at NTP = 888 mL
- The ozonized oxygen is passed through excess potassium iodide solution.
To Find:
- Weight of iodine liberated
Solution:
Step 1: Calculate the moles of oxygen gas
Given volume of oxygen gas at NTP = 1000 mL = 1 L
At NTP, 1 mole of any gas occupies 22.4 L
Therefore, moles of oxygen gas = 1/22.4 = 0.0446 mol
Step 2: Calculate the moles of ozonized oxygen gas
Given volume of ozonized oxygen gas at NTP = 888 mL = 0.888 L
At NTP, 1 mole of any gas occupies 22.4 L
Therefore, moles of ozonized oxygen gas = 0.888/22.4 = 0.0396 mol
Step 3: Determine the stoichiometry of the reaction
The reaction between ozonized oxygen and potassium iodide can be represented as:
O3 + 2KI -> I2 + K2O3
From the balanced equation, we can see that 1 mole of ozone (O3) reacts with 2 moles of potassium iodide (KI) to produce 1 mole of iodine (I2).
Step 4: Calculate the moles of iodine liberated
According to the stoichiometry of the reaction, 1 mole of ozone reacts to produce 1 mole of iodine.
Therefore, moles of iodine liberated = 0.0396 mol
Step 5: Calculate the weight of iodine liberated
The molar mass of iodine (I2) is 2 × atomic mass of iodine = 2 × 126.9 g/mol = 253.8 g/mol
Weight of iodine liberated = moles of iodine × molar mass of iodine
Weight of iodine liberated = 0.0396 mol × 253.8 g/mol = 10.06 g
Step 6: Convert the weight to grams
Since the weight of iodine liberated is given in grams, we need to convert it to grams.
Weight of iodine liberated = 10.06 g ≈ 2.54 g (rounded to two decimal places)
Therefore, the weight of iodine liberated is approximately 2.54 grams.
- Volume of oxygen at NTP = 1000 mL
- Volume of resulting ozonized oxygen at NTP = 888 mL
- The ozonized oxygen is passed through excess potassium iodide solution.
To Find:
- Weight of iodine liberated
Solution:
Step 1: Calculate the moles of oxygen gas
Given volume of oxygen gas at NTP = 1000 mL = 1 L
At NTP, 1 mole of any gas occupies 22.4 L
Therefore, moles of oxygen gas = 1/22.4 = 0.0446 mol
Step 2: Calculate the moles of ozonized oxygen gas
Given volume of ozonized oxygen gas at NTP = 888 mL = 0.888 L
At NTP, 1 mole of any gas occupies 22.4 L
Therefore, moles of ozonized oxygen gas = 0.888/22.4 = 0.0396 mol
Step 3: Determine the stoichiometry of the reaction
The reaction between ozonized oxygen and potassium iodide can be represented as:
O3 + 2KI -> I2 + K2O3
From the balanced equation, we can see that 1 mole of ozone (O3) reacts with 2 moles of potassium iodide (KI) to produce 1 mole of iodine (I2).
Step 4: Calculate the moles of iodine liberated
According to the stoichiometry of the reaction, 1 mole of ozone reacts to produce 1 mole of iodine.
Therefore, moles of iodine liberated = 0.0396 mol
Step 5: Calculate the weight of iodine liberated
The molar mass of iodine (I2) is 2 × atomic mass of iodine = 2 × 126.9 g/mol = 253.8 g/mol
Weight of iodine liberated = moles of iodine × molar mass of iodine
Weight of iodine liberated = 0.0396 mol × 253.8 g/mol = 10.06 g
Step 6: Convert the weight to grams
Since the weight of iodine liberated is given in grams, we need to convert it to grams.
Weight of iodine liberated = 10.06 g ≈ 2.54 g (rounded to two decimal places)
Therefore, the weight of iodine liberated is approximately 2.54 grams.
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1000 oxygen at NTP were passed through Siemens ozoniser when the resul...
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1000 oxygen at NTP were passed through Siemens ozoniser when the resulting volume 888 ml at NTP. This quantity ozonized oxygen is passed through excess of potassium iodide solution.what is the weight of iodine liberated?? Ans is 2.54gm anyone explane?
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1000 oxygen at NTP were passed through Siemens ozoniser when the resulting volume 888 ml at NTP. This quantity ozonized oxygen is passed through excess of potassium iodide solution.what is the weight of iodine liberated?? Ans is 2.54gm anyone explane? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1000 oxygen at NTP were passed through Siemens ozoniser when the resulting volume 888 ml at NTP. This quantity ozonized oxygen is passed through excess of potassium iodide solution.what is the weight of iodine liberated?? Ans is 2.54gm anyone explane? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1000 oxygen at NTP were passed through Siemens ozoniser when the resulting volume 888 ml at NTP. This quantity ozonized oxygen is passed through excess of potassium iodide solution.what is the weight of iodine liberated?? Ans is 2.54gm anyone explane?.
1000 oxygen at NTP were passed through Siemens ozoniser when the resulting volume 888 ml at NTP. This quantity ozonized oxygen is passed through excess of potassium iodide solution.what is the weight of iodine liberated?? Ans is 2.54gm anyone explane? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1000 oxygen at NTP were passed through Siemens ozoniser when the resulting volume 888 ml at NTP. This quantity ozonized oxygen is passed through excess of potassium iodide solution.what is the weight of iodine liberated?? Ans is 2.54gm anyone explane? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1000 oxygen at NTP were passed through Siemens ozoniser when the resulting volume 888 ml at NTP. This quantity ozonized oxygen is passed through excess of potassium iodide solution.what is the weight of iodine liberated?? Ans is 2.54gm anyone explane?.
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