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Get Full Access to Principles Of General Chemistry - 2 Edition - Chapter 3 - Problem 50p
Get Full Access to Principles Of General Chemistry - 2 Edition - Chapter 3 - Problem 50p

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# Calculate the maximum numbers of moles and grams of iodic ISBN: 9780073511085 71

## Solution for problem 50P Chapter 3

Principles of General Chemistry | 2nd Edition

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Problem 50P Problem 50P

Calculate the maximum numbers of moles and grams of iodic acid (HIO3) that can form when 635 g of iodine trichloride reacts with 118.5 g of water: What mass of the excess reactant remains?

Step-by-Step Solution:

### Solution 50P

Here we have to calculate the maximum numbers of moles and grams of iodic acid (HIO3) that can form when 635 g of iodine trichloride reacts with 118.5 g of water and have to find out the mass of the excess reactant remains present in the reaction.

The given unbalanced chemical equation is,

ICl3 + H2O → ICl + HIO3 + HCl

Step 1

Given:

Mass of ICl3 = 6 35g

Molar mass of ICl3 = 233.2 g

Mass of H2O = 118.5 g

Molar mass of H2O = 18.02 g

Molar mass of HIO3 = 175.9 g

1st we have to write a balanced chemical equation

2ICl3 + 3H2O → ICl + HIO3 + 5HCl

In order to calculate the maximum numbers of moles and grams of iodic acid (HIO3) formed from ICl3 we have to find out the limiting reactant.

Moles of HIO3 formed from ICl3 :

In the above reaction 1 mole of HIO3 has formed from 2 moles of ICl3 , then the mole of HIO3 can be calculated as,

Mass of ICl3  = 685 g ICl3  = 1.4686 or 1.47 mol of HIO3

Moles of HIO3 formed from H2O :

In the above reaction 1 mole of HIO3 has formed from 3 moles of H2O , then the mole of HIO3 can be calculated as,

Mass of H2O  = 118.5 g H2O  = 2.17166 or 2.17 mol of HIO3

Thus from the above calculation we can say that ICl3 is the limiting reactant because it produces less amount of HIO3 i.e 1.47 mol.

Step 2 of 3

Step 3 of 3

##### ISBN: 9780073511085

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