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4 years ago

8

Which is not a compound?Electrons spin in shells around the nucleus. The closest shell (n = 1) can contain a maximum of _____ el

ectrons?

1 answer:

melisa1 [442]4 years ago

6 0

The closest shell (n = 1) can contain a maximum of 2 electrons.

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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

poizon [28]

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

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 = $44.9g H_2O$

To learn more about Chemical Reaction

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2 years ago

Which statement is true if the mc001-1.jpgHrxn of a reaction is positive? The reaction is endothermic. The reaction is exothermi

Bas_tet [7]

The answer is A.( The reaction is endothermic) I had it on my test and got it right

4 0

3 years ago

Read 2 more answers

2k+Cl2=2KCI how many grams of KCl is produced from 8.50 g of K and excess Cl2

ZanzabumX [31]

Answer:

Mass = 16.4 g

Explanation:

Given data:

Mass of K = 8.50 g

Mass of KCl produced = ?

Solution:

Chemical equation:

2K + Cl₂ → 2KCl

Number of moles of K:

Number of moles = mass/ molar mass

Number of moles = 8.50 g/ 39 g/mol

Number of moles = 0.22 mol

Now we will compare the moles of potassium and potassium chloride.

K : KCl

2 : 2

0.22 : 0.22

Mass of KCl:

Mass = number of moles × molar mass

Mass = 0.22 mol × 74.55 g/mol

Mass = 16.4 g

4 0

3 years ago

How many years have the earth

Ivenika [448]

We need more information to answer this.

6 0

3 years ago

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2 AgNO3(aq) + CaCl2(aq) -----> 2 AgCl(s) + Ca(NO3)2(aq)

yan [13]

5.732 grams of AgCl is formed when 0.200 L of 0.200 M AGNO3 reacts with an excess of CaCl2.

Explanation:

The balanced equation:

2 AgNO3(aq) + CaCl2(aq) -----> 2 AgCl(s) + Ca(NO3)2(aq)

data given:

volume of AgNO3 = 0.2 L

molarity of AgNO3 = 0.200 M

atomic weight of AgCl= 143.32 gram/mole

from the formula, number of moles can be calculated

Molarity = $\frac{number of moles}{volume in litres}$

number of moles of AgNO3 = 0.04

From the reaction:

2 moles of AgNO3 reacts to form 2 moles of AgCl

0.04 moles of AgNO3 reacts to form x mole of AgCl

$\frac{2}{2}$ = $\frac{x}{0.04}$

= 0.04 moles of AgCl is formed

mass of AgCl formed is calculated by multiplying number of moles with atomic mass of AgCl

mass of AgCl = 0.04 x 143.32

= 5.732 grams of AgCl is formed.

4 0

3 years ago