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N76 [4]
4 years ago
9

When aluminum, AlAl, metal is dipped in an aqueous solution of hydrochloric acid, HClHCl, hydrogen gas, H2H2, is produced with t

he formation of an aluminum chloride, AlCl3AlCl3, solution. Write the balanced chemical equation showing the phases of reactants and products.
Chemistry
1 answer:
kifflom [539]4 years ago
5 0

Answer:

Balanced equation: 2Al(s)+6HCl(aq.)\rightarrow 2AlCl_{3}(aq.)+3H_{2}(g)

Explanation:

  • Al is an metallic species with a standard reduction potential of -1.66 V( E_{Al^{3+}\mid Al}^{0}=-1.66V).
  • H_{2} is present at the intermediate position in electrochemical series with a standard reduction potential of 0 V (E_{H^{+}\mid H_{2}}^{0}=0V).
  • So, when Al is dipped in aqueous solution of HCl, Al is readily oxidized to produce Al^{3+} whereas H^{+} is reduced to H_{2} .
  • Hence, as a whole, aqueous AlCl_{3} and gaseous H_{2} is produced as products.

Balanced equation: 2Al(s)+6HCl(aq.)\rightarrow 2AlCl_{3}(aq.)+3H_{2}(g)

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Answer:

Explanation:Explanation:

Your starting point here will be the ideal gas law equation

∣

∣

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¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

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∣

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0.0821

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⋅

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Now, you will have to manipulate this equation in order to find a relationship between the density of the gas,

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M

M

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You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass

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∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

M

M

=

m

n

a

a

∣

∣

−−−−−−−−−−−−−

(

1

)

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass

m

of this gas, you can express its density as the ratio between

m

and the volume it occupies

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

ρ

=

m

V

a

a

∣

∣

−−−−−−−−−−−

(

2

)

Plug equation

(

1

)

into the ideal gas law equation to get

P

V

=

m

M

M

⋅

R

T

Rearrange to get

P

V

⋅

M

M

=

m

⋅

R

T

P

⋅

M

M

=

m

V

⋅

R

T

M

M

=

m

V

⋅

R

T

P

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(

2

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to write

M

M

=

ρ

⋅

R

T

P

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⋅

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K

⋅

(

273.15

+

37

)

K

0.990

atm

M

M

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

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−

1

a

a

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∣

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<em />

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