Calculate the maximum numbers of moles and grams of iodic acid (HIO₃) that can form when 635 g of iodine trichloride reacts with

Question

solong [7]

2 years ago

8

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:
ICl₃ + H₂O → ICl + HIO₃ + HCl [unbalanced]
How many grams of the excess reactant remains?

Chemistry

1 answer:

poizon [28] · Answer 1 · 2023-05-07T11:36:18+03:00

What is Chemical Reaction?

A chemical reaction occurs when a certain group of molecules converts into another form without affecting their nuclei; only the transfer or sharing of electrons and the building and breaking of bonds occur.

Main Content

Known :

$\mathrm{ICl}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ICl}+\mathrm{HIO}_{3}+\mathrm{HCl}$

Mass of $ICI_3 = 635g$

Mass of $H_2O = 118.5g$

Calculations :

First, balance the given chemical equation by place 2,3 and 5 as the coefficients of $ICI_3, H_2O and HCl,\\$ Respectively

$2 \mathrm{ICl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ICl}+\mathrm{HIO}_{3}+5 \mathrm{HCl}$

We multiply the given mass of $ICl_3$ by the reciprocal of its molar mass to get the number of moles. The molar mass of $ICl_3 is 233.26g/mol$

$\text { Moles of } \mathrm{ICl}_{3}=635 \mathrm{~g} \mathrm{ICl}_{3} \times \frac{1 \mathrm{~mol} \mathrm{ICl}_{3}}{233.26 \mathrm{~g} \mathrm{ICl}_{3}}=2.7223 \mathrm{~mol} \mathrm{ICl}_{3}$

Them, we multiply the ratio between $ICl_3 and HIO_3$ . based on the chemical equation, the molar ratio 1 mol $HIO_3/2 mol ICl_3$ .

$\text { Moles of } \mathrm{HIO}_{3} \text { formed }=2.7223 \mathrm{~mol} \mathrm{ICl}_{3} \times \frac{1 \mathrm{~mol} \mathrm{HIO}_{3}}{2 \mathrm{~mol} \mathrm{ICl}_{3}}=1.3612 \mathrm{~mol} \mathrm{HIO}_{3}$

For water, we again multiply the given mass $H_2O$ by the reciprocal of its molar mass to get the number if moles. The molar mass of $H_2O$ is 18.02g/mol

$\text { Moles of } \mathrm{H}_{2} \mathrm{O}=118.5 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} \times \frac{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{18.02 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}}=6.5760 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}$

Then, we multiply the molar ratio between H2O and HIOs. Based on the chemical equation, the molar ratio is 1 mol $HIO_3/3 Mole H_2O$

$\text { Moles of } \mathrm{HIO}_{3} \text { formed }=6.5760 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O} \times \frac{1 \mathrm{~mol} \mathrm{HIO}_{3}}{3 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}=2.1920 \mathrm{~mol} \mathrm{HIO}_{3}$

We can see that the limiting reactant is $ICL_3$ since the given mass of $ICl_3$ forms lesser product than water does. Thus , the maximum number of moles of $HIO_3$ formed is 1.36 mol $HIO_3$ . We now multiply the molar mas of $HIO_3$ to the calculated number of moles. The molar mass of $HIO_3$ is 175.91 g/mol.

Mass of $HIO_3$ formed (Max) $=1.3612 \mathrm{~mol} \mathrm{HIO}_{3} \times \frac{175.91 \mathrm{~g} \mathrm{HIO}_{3}}{1 \mathrm{~mol} \mathrm{HIO}}=239 \mathrm{~g} \mathrm{HIO}_{3}$

We multiply the number of moles of $ICl_3$ by the molar ratio between $ICl_3$ and $H_2O$ which is $3 mol H_2O mol ICl_3$ we get the number of moles of $H_2O$ reacted. Then, we multiply the molar mass of water.

Mass of $H_2O$ reacted $=2.7223 \mathrm{~mol} \mathrm{ICl} 3 \times \frac{3 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{2 \mathrm{~mol} \mathrm{ICl}_{3}} \times \frac{18.02 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}$