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lana [24]
3 years ago
5

Read the following chemical equations.

Chemistry
1 answer:
Viktor [21]3 years ago
4 0

Answer:

D. Chlorine is reduced in reaction 1 and iron is oxidized in reaction 2.

Step-by-step explanation:

One way to determine if a substance has been oxidized or reduced is to calculate its <em>oxidation number</em>.

The important rules for oxidation numbers are:

  1. The oxidation number of an element is zero.
  2. The oxidation number of H is usually +1.
  3. The oxidation number of Cl is usually -1.
  4. The sum of all the oxidation numbers in a compound is zero.

Also,

  • An increase in oxidation number is oxidation.
  • A decrease in oxidation number is reduction.

1. \stackrel{\hbox{+1}}{\hbox{H}_{2}}\stackrel{\hbox{-2}}{\hbox{S}} + \stackrel{\hbox{0}}{\hbox{Cl}}_{2}  \longrightarrow\stackrel{\hbox{+1}}{\hbox{ 2H}}\stackrel{\hbox{-1}}{\hbox{Cl}} +\stackrel{\hbox{0}}{\hbox{S}}

2. \stackrel{\hbox{0}}{\hbox{Fe}} + \stackrel{\hbox{0}}{\hbox{S}} \longrightarrow \stackrel{\hbox{+2}}{\hbox{ Fe}}\stackrel{\hbox{-2}}{\hbox{S}}

Thus, the oxidation number of S in H₂S and FeS is +2.

In Reaction 1, the oxidation number of Cl decreases from 0 to -1. The Cl is reduced.

In Reaction 2, the oxidation number of Fe changes from 0 to +2. The Fe is oxidized.

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Calculate the number of moles of Cl2 produced at equilibrium when 3.98 mol of PCl5 is heated at 283.9 deg celsius in a vessel ha
ddd [48]

Answer:

1.274 moles

Explanation:

The equation for the reaction can be represented as follows:

                       PCl_5(g)           ⇄       PCl_3(g)     +         Cl_2(g)

K = 0.060

K = \frac{[PCl_3][Cl_2]}{[PCl_5]}

Concentration of   PCl_5(g)  = \frac{numbers of moles}{volume}

Concentration of   PCl_5(g)  = \frac{3.98}{10.0}

Concentration of   PCl_5(g)  = 0.398 moles

If we construct an ICE table for the above equation; we have:

                       PCl_5(g)           ⇄       PCl_3(g)     +         Cl_2(g)

Initial                0.398                          0                          0

Change            - x                               + x                        + x

Equilibrium    (0.398 - x)                      x                          x

K = \frac{[PCl_3][Cl_2]}{[PCl_5]}

K = \frac{[x][x]}{[0.398-x]}

K = \frac{x^2}{0.398-x}

0.060 = \frac{x^2}{0.398-x}

0.06(0.398-x) = x²

0.02388 - 0.060x = x²

x² + 0.060x - 0.02388 = 0               (quadratic equation)

a = 1;       b= 0.06;      c= -0.02388

Using quadratic formula;

=  \frac{-b+/-\sqrt{b^2-4ac} }{2a}

= \frac{-0.06+/-\sqrt{(0.06)^2-4(1)(-0.02388)} }{(2*1)}

= \frac{-0.060+/-\sqrt{0.0036+0.09552} }{2}

= \frac{-0.06+/-\sqrt{0.09912} }{2}

= \frac{-0.06+/-0.3148}{2}

= \frac{-0.060+0.3148}{2}   or \frac{-0.060-0.3148}{2}

= \frac{0.2548}{2}  or \frac{-0.3748}{2}

= 0.1274 or -0.1874

We go by the positive value which says:

[x] = 0.1274 M

number of moles = 0.1274 × 10.0

= 1.274 moles

∴ the number of moles  of Cl₂ produced at equilibrium = 1.274 moles

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0.1 M

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