Respuesta :
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 :
[tex]\mathrm{ICl}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ICl}+\mathrm{HIO}_{3}+\mathrm{HCl}[/tex]
Mass of [tex]ICI_3 = 635g[/tex]
Mass of [tex]H_2O = 118.5g[/tex]
Calculations :
First, balance the given chemical equation by place 2,3 and 5 as the coefficients of [tex]ICI_3, H_2O and HCl,\\[/tex] Respectively
[tex]2 \mathrm{ICl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ICl}+\mathrm{HIO}_{3}+5 \mathrm{HCl}[/tex]
We multiply the given mass of [tex]ICl_3[/tex] by the reciprocal of its molar mass to get the number of moles. The molar mass of [tex]ICl_3 is 233.26g/mol[/tex]
[tex]\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}[/tex]
Them, we multiply the ratio between [tex]ICl_3 and HIO_3[/tex]. based on the chemical equation, the molar ratio 1 mol [tex]HIO_3/2 mol ICl_3[/tex].
[tex]\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}[/tex]
For water, we again multiply the given mass [tex]H_2O[/tex] by the reciprocal of its molar mass to get the number if moles. The molar mass of [tex]H_2O[/tex] is 18.02g/mol
[tex]\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}[/tex]
Then, we multiply the molar ratio between H2O and HIOs. Based on the chemical equation, the molar ratio is 1 mol [tex]HIO_3/3 Mole H_2O[/tex]
[tex]\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}[/tex]
We can see that the limiting reactant is [tex]ICL_3[/tex] since the given mass of [tex]ICl_3[/tex] forms lesser product than water does. Thus , the maximum number of moles of [tex]HIO_3[/tex] formed is 1.36 mol [tex]HIO_3[/tex]. We now multiply the molar mas of [tex]HIO_3[/tex] to the calculated number of moles. The molar mass of [tex]HIO_3[/tex] is 175.91 g/mol.
Mass of [tex]HIO_3[/tex] formed (Max) [tex]=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}[/tex]
We multiply the number of moles of [tex]ICl_3[/tex] by the molar ratio between [tex]ICl_3[/tex] and [tex]H_2O[/tex] which is [tex]3 mol H_2O mol ICl_3[/tex] we get the number of moles of [tex]H_2O[/tex] reacted. Then, we multiply the molar mass of water.
Mass of [tex]H_2O[/tex] reacted [tex]=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}}[/tex]
Mass of [tex]H_2O[/tex] reacted = 73.6g [tex]H_2O[/tex]
We subtract the mass of [tex]H_2O[/tex] reacted from the given mass of [tex]H_2O[/tex].
Mass of [tex]H_2O[/tex] = 118.5 - 73.6g = [tex]44.9g H_2O[/tex]
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