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3 years ago

SnCl4 → Sn + __ Cl2

Chemistry

1 answer:

Kisachek [45]3 years ago

6 0

Answer:

Sn (s) + 2 Cl2 (g) → SnCl4 (l)

This is an oxidation-reduction (redox) reaction:

4 Cl0 + 4 e- → 4 Cl-I

(reduction)

Sn0 - 4 e- → SnIV

(oxidation)

Cl2 is an oxidizing agent, Sn is a reducing agent.

Reactants:

Names: Tin source: wikidata, accessed: 2019-09-07source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, Sn source: wikidata, accessed: 2019-09-07, Element 50 source: wikidata, accessed: 2019-09-07

Appearance: White crystalline powder source: ICSC, accessed: 2019-09-04; Gray to almost silver-white, ductile, malleable, lustrous solid. source: NIOSH NPG, accessed: 2019-09-02

Cl2

Names: Chlorine source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, Molecular chlorine source: NIOSH NPG, accessed: 2019-09-02

Appearance: Greenish-yellow compressed liquefied gas with pungent odour source: ICSC, accessed: 2019-09-04; Greenish-yellow gas with a pungent, irritating odor. [Note: Shipped as a liquefied compressed gas.] source: NIOSH NPG, accessed: 2019-09-02

Products:

SnCl4 – Tetrachlorostannane source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02, Tin tetrachloride source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04, Tin(IV) chloride source: wikipedia, accessed: 2019-09-28

Other names: Stannic chloride source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04, Tin(iv) chloride (anhydrous) source: ICSC, accessed: 2019-09-04

Appearance: Colorless to slightly yellow fuming liquid source: wikipedia, accessed: 2019-09-28; Colourless or slightly yellow fuming liquid with pungent odour source: ICSC, accessed: 2019-09-04

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$