How many moles of Fluorine (F2) are needed to completely react 8.0 moles of NF3?

Chemistry

1 answer:

Anna [14]3 years ago

8 0

Answer:

12 moles of F₂

Explanation:

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

N₂ + 3F₂ —> 2NF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Finally, we shall determine the number of mole of F₂ needed to produce 8 moles of NF₃. This can be obtained as illustrated below:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Therefore, Xmol of F₂ will react to produce 8 moles of NF₃ i.e

Xmol of F₂ = (3 × 8)/2

Xmol of F₂ = 12 moles

Thus, 12 moles of F₂ is needed for the reaction.

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A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$