Question

Cobalt(II) chloride reacts with fluorine in a single replacement reaction to produce cobalt(II) fluoride and chlorine gas. How many grams of fluorine are required to produce 124.13 g of cobalt(II)fluoride?

Answer 1

The balanced equation of the reaction is:

CoCl2 + F2 → CoF2 + Cl2

Now, we will follow the next steps to solve the question.

1. We find the moles present in 124.13 g of cobalt(II)fluoride (CoF2) using the molar molar mass of CoF2. The molar mass of CoF2 is: 96.93g/mol

2. By stoichiometry we find the moles of fluorine (F2) needed. Since the ratio CoF2 to F2 is 1, the moles will be the same as those produced from cobalt(II)fluoride.

3. We find the grams of fluorine by multiplying the moles by the molar mass of fluorine. The molar mass of fluorine is 38.00 g/mol

Let's proceed with the calculations:

1. Moles of CoF2

`\begin{gathered} molCoF_2=givengCoF_2*(1molCoF_2)/(MolarMass,gCoF_2) \\ molCoF_2=124.13gCoF_2*(1molCoF_2)/(96.93gCoF_2)=1.28molCoF_2 \end{gathered}`

2. Moles of F2

`molF_2=1.28molCoF_2*(1molF_2)/(1molCoF_2)=1.28molF_2`

3. Grams of F2

`\begin{gathered} gF_2=givenmolF_2*(MolarMass,gF_2)/(1molF_2) \\ gF_2=1.28molF_2*(38.00gF_2)/(1molF_2)=48.66gF_2 \end{gathered}`

Answer: To produce 124.13 grams of cobalt(II)fluoride are required 48.66grams of fluorine