Answer to Question #181538 in Chemistry for angela nicole udan

Question

If 0.200 grams CCl4 is mixed with 0.175 grams of SbF3, how many grams of CCl2F2 is formed

Answer 1

Solution:

Balanced chemical equation:

3CCl₄ + 2SbF₃ → 3CCl₂F₂ + 2SbCl₃

According to the equation above: n(CCl₄)/3 = n(SbF₃)/2 = n(CCl₂F₂)/3

The molar mass of CCl₄ is 153.82 g mol^-1.

The molar mass of SbF₃ is 178.75 g mol^-1.

The molar mass of CCl₂F₂ is 120.91 g mol^-1.

Determine moles of each reactant:

0.200 g CCl₄ × (1 mol CCl₄ / 153.82 g CCl₄) = 0.001300 mol CCl₄

0.175 g SbF₃ × (1 mol SbF₃ / 178.75 g SbF₃) = 0.000979 mol SbF₃

Choose one reactant and determine how many moles of the other reactant are necessary to completely react with it. Let's choose CCl₄:

n(SbF₃) = 2 × n(CCl₄) / 3 = (2 × 0.001300 mol) / 3 = 0.000867 mol

 

The calculation above means that we need 0.000867 mol of SbF₃ to completely react with CCl₄.

We have 0.000979 mol of SbF₃ and therefore more than enough antimony trifluoride.

Thus antimony trifluoride (SbF₃) is in excess and carbon tetrachloride (CCl₄) is the limiting reagent.

Hence,

n(CCl₄)/3 = n(CCl₂F₂)/3

n(CCl₄) = n(CCl₂F₂)

n(CCl₂F₂) = n(CCl₄) = 0.001300 mol

(0.001300 mol CCl₂F₂) × (120.91 g CCl₂F₂ / 1 mol CCl₂F₂) = 0.157183 g CCl₂F₂ = 0.157 g CCl₂F₂

Answer: 0.157 grams of CCl₂F₂ is formed.