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Harrizon [31]
3 years ago
9

Halogenation of propane​

Chemistry
1 answer:
aniked [119]3 years ago
5 0

Answer:

The halogenation of propane discloses an interesting feature of these reactions. All the hydrogens in a complex alkane do not exhibit equal reactivity. For example, propane has eight hydrogens, six of them being structurally equivalent primary, and the other two being secondary

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CF4 + Br2 CBr4 + F2
luda_lava [24]

Answer:

Answers are in the explanation

Explanation:

Based on the reaction:

CF₄ + 2Br₂ → CBr₄ + 2F₂

The mole ratio of CF₄ is:

CF₄:Br₂ = 1:2

CF₄:CBr₄ = 1:1

CF₄:F₂ = 1:2

<em>Moles F2:</em>

Molar mass CF₄: 88.0g/mol

57.0g * (1mol / 88.0g) = 0.6477 moles CF₄ * (2mol F₂ / 1mol CBr₄) =

<h3>1.30 moles F₂</h3><h3 />

<em>Mass Br2:</em>

Molar mass CBr₄: 331.63g/mol

250.0g * (1mol / 331.63g) = 0.7539 moles CBr₄ * (2mol Br₂ / 1mol CF₄) =

1.51 moles Br₂ * (159.808g / mol) =

<h3>241g Br2</h3><h3 /><h3 />

<em>Moles F2:</em>

4.8 moles CF₄ * (2mol F₂ / 1mol CF₄) =

<h3>9.6 moles F₂</h3><h3 />

<em />

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