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

0.73moles

128.48‬g

Step-by-step 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 =
`(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 =
`(341)/(233.5)` = 1.46moles

Number of moles of H₂O =
`(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
`(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
`(1.46)/(2)` = 0.73moles

Mass of HIO₃ = number of moles x molar mass

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

= 128.48‬g