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Vinil7 [7]
3 years ago
15

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

ving a capacity of 10.0L. At 283.9, K = 0.060 for this dissociation.
Chemistry
1 answer:
ddd [48]3 years ago
7 0

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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3 years ago
How many grams of carbon dioxide will form if 5.5 g of C3H8 burns in 15 g of O2?
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Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
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3 years ago
What particle decay is this? 210 83 Bi→210 84 Po ​
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Answer:

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Explanation:

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How many grams of ammonia (NH3) can be produced by the synthesis of excess hydrogen gas (H2) and 253.8 grams of nitrogen gas (N2
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Answer:

308.2 g of NH₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3H₂ + N₂ —> 2NH₃

Next, we shall determine the mass of N₂ that reacted and the mass of NH₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of N₂ = 2 × 14 = 28 g/mol

Mass of N₂ from the balanced equation = 1 × 28 = 28 g

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Mass of NH₃ from the balanced equation = 2 × 17 = 34 g

Summary:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Finally, we shall determine the mass of NH₃ produced by the reaction of 253.8 g of N₂. This can be obtained as illustrated below:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Therefore, 253.8 g of N₂ will react to produce = (253.8 × 34)/28 = 308.2 g of NH₃.

Thus, 308.2 g of NH₃ were obtained from the reaction.

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3 years ago
What are two qualities of metals?​ Describe each
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Answer:Typical physical properties of metals : high melting points. good conductors of electricity. good conductors of heat.

Explanation:

4 0
1 year ago
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