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3 years ago

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Consider four different samples: aqueous LiBr, molten LiBr, aqueous LiF, and molten LiF. Current run through each sample produce

s one of the following products at the anode: liquid bromine, fluorine gas, or oxygen gas. Match each sample to its anodic product from aqueous LiBr, Molten LiBr, aqueous LiF, and molten LiFA. Liquid bromine,
B. Fluorine gas,
C. oxygen gas

1 answer:

Verizon [17]3 years ago

5 0

Answer:

Following are the solution to the given question:

Explanation:

Please find the matching in attached file.

During the electrolysis of Molten LiBr : Li is reduced and Br are oxidized .

Lithium Metal is produced at the Cathod during the electrolysis of Molten LiBR .

In the aquous LiBr : In aquous LiBr potential of Li is greater than the of water then Li is reduced to produce solid Li ion. As well As aqueous LiF ( electrolysis)and Molten LiF electrolysis produce the gas.

aqueous LiF: Oxygen gas.

Molten LiF : Flourine gas.

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

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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 $HIO_3/3 Mole H_2O$

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

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

Mass of $H_2O$ reacted = 73.6g $H_2O$

We subtract the mass of $H_2O$ reacted from the given mass of $H_2O$ .

Mass of $H_2O$ = 118.5 - 73.6g = $44.9g H_2O$

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Answer is: 7.8 lb of 21% aluminum and 33.2 ib of <span>42% aluminum.</span>

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ω</span>₃<span> = 38% ÷ 100% = 0.38.
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m₂<span> = ?.
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m₁ = 41 lb - m₂<span>.
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