Question

Calculate the maximum numbers of moles and grams of iodic acid (HIO3) that can form when 341 g of iodine trichloride reacts with 151.7 g of water:
ICl3 + H2O → ICl + HIO3 + HCl [unbalanced]

Answer 1

Answer:

0.73moles

128.48‬g

Explanation:

Given parameters:

Mass of ICl₃  = 341g

Mass of H₂O = 151.7g

Unknown:

Maximum numbers of moles and grams of iodic acid = ?

Solution:

To solve this problem, one needs to obtain a balanced chemical equation for a start;

                   2ICl₃   +    3H₂O    →   ICI    +     HIO₃   +   5HCl

Now, we need to find the limiting reagent because it determines the extent of the reaction;

         Number of moles  = $\frac{mass}{molar mass}$

molar mass of ICl₃ = 127 + 3(35.5) = 233.5g/mol

molar mass of H₂O  = 2(1) + 16  = 18g/mol

Number of moles of ICl₃ = $\frac{341}{233.5}$ = 1.46moles

Number of moles of H₂O  = $\frac{151.7}{18}$  = 8.43moles

 From the balanced equation;

               2 moles of ICl₃ combined with 3 moles of water

             1.46 moles of  ICl₃ will need $\frac{1.46 x 3}{2}$  = 2.19 moles

But we were given 8.43moles of water;

           This suggests that water is excess

Now we can relate the known to the unknown:

              2 moles of  ICl₃  produced 1 mole of  HIO₃

    then  1.46 mole of  ICl₃  will produce $\frac{1.46}{2}$  = 0.73moles

Mass of HIO₃ = number of moles x molar mass

                       = 0.73moles  x (1 + 127 + 3(16))

                        = 128.48‬g