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What weight of HNO3 is needed to convert 5g of Iodine into Iodic acid acc. To the reaction
I2 + HNO3→ HIO3 + NO2 +H2O
I2 + HNO3→ HIO3 + NO2 +H2O
- a)12.4g
- b)24.8g
- c)0.248g
- d)49.6g
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
What weight of HNO3 is needed to convert 5g of Iodine into Iodic acid ...
The balanced reaction is:
I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O
127*2g of iodine requires 63*10g of HNO3
Thus, 5g of HNO requires:
63*10/127*2 *5
= 12.g
Most Upvoted Answer
What weight of HNO3 is needed to convert 5g of Iodine into Iodic acid ...
Given:
Mass of Iodine (I2) = 5g
Reaction:
I2 + HNO3 → HIO3 + NO2 + H2O
To find:
Weight of HNO3 needed to convert 5g of Iodine into Iodic acid (HIO3)
Solution:
The balanced chemical equation for the reaction shows that one mole of Iodine reacts with one mole of Nitric acid (HNO3) to form one mole of Iodic acid (HIO3), Nitrogen dioxide (NO2) and water (H2O).
From the balanced equation, we can calculate the number of moles of Iodine (I2) present in 5g of it.
Molar mass of Iodine (I2) = 2 x 126.9 = 253.8g/mol (2 atoms of Iodine in 1 molecule)
Number of moles of Iodine (I2) = Mass/Molar mass = 5/253.8 = 0.0197 mol
Since the reaction is 1:1 between Iodine and Nitric acid, the number of moles of Nitric acid (HNO3) required to react with 0.0197 mol of Iodine is also 0.0197 mol.
Now, from the balanced equation, we know that 1 mole of HNO3 reacts with 1 mole of I2 to produce 1 mole of HIO3. Therefore, the number of moles of HNO3 required to react with 0.0197 mol of Iodine is also 0.0197 mol.
To calculate the weight of HNO3 required, we need to use its molar mass.
Molar mass of HNO3 = 1 + 14 + 48 = 63g/mol
Weight of HNO3 required = Number of moles x Molar mass
= 0.0197 x 63
= 1.24g
Therefore, the weight of HNO3 required to convert 5g of Iodine into Iodic acid is 1.24g or approximately 12.4g (option A).
Mass of Iodine (I2) = 5g
Reaction:
I2 + HNO3 → HIO3 + NO2 + H2O
To find:
Weight of HNO3 needed to convert 5g of Iodine into Iodic acid (HIO3)
Solution:
The balanced chemical equation for the reaction shows that one mole of Iodine reacts with one mole of Nitric acid (HNO3) to form one mole of Iodic acid (HIO3), Nitrogen dioxide (NO2) and water (H2O).
From the balanced equation, we can calculate the number of moles of Iodine (I2) present in 5g of it.
Molar mass of Iodine (I2) = 2 x 126.9 = 253.8g/mol (2 atoms of Iodine in 1 molecule)
Number of moles of Iodine (I2) = Mass/Molar mass = 5/253.8 = 0.0197 mol
Since the reaction is 1:1 between Iodine and Nitric acid, the number of moles of Nitric acid (HNO3) required to react with 0.0197 mol of Iodine is also 0.0197 mol.
Now, from the balanced equation, we know that 1 mole of HNO3 reacts with 1 mole of I2 to produce 1 mole of HIO3. Therefore, the number of moles of HNO3 required to react with 0.0197 mol of Iodine is also 0.0197 mol.
To calculate the weight of HNO3 required, we need to use its molar mass.
Molar mass of HNO3 = 1 + 14 + 48 = 63g/mol
Weight of HNO3 required = Number of moles x Molar mass
= 0.0197 x 63
= 1.24g
Therefore, the weight of HNO3 required to convert 5g of Iodine into Iodic acid is 1.24g or approximately 12.4g (option A).
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Community Answer
What weight of HNO3 is needed to convert 5g of Iodine into Iodic acid ...
ccording to the choices given, the mass of iodine should be 50g.
CALCULATIONS :
From the given equation, the mole ratio is 1 : 1
Molar mass of iodine gas = (127 X 2) g / mol = 254 g / mol
Moles = 50 / 254 = 0.19685 moles
Moles of iodine = moles of nitric acid = 0.19685 moles.
Molar mass of nitric acid = 63g / mol
Mass of nitric acid = 63 X 0.19685 = 12.40 g
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