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Sunny_sXe [5.5K]
3 years ago
13

How many moles are in 55.98g CF2Cl2

Chemistry
2 answers:
Karolina [17]3 years ago
7 0
No of moles = given mass ÷ molecular mass
n = 55.98 ÷ (12+19×2+35.5×2)
Nataliya [291]3 years ago
6 0

Explanation:

Number of moles are defined as the mass divided by molar mass.

Mathematically,       No. of moles = \frac{mass}{\text{molar mass}}

It is given that mass is 55.98 g and molar mass of CF_{2}Cl_{2} is 120.91 g/mol.

Therefore, calculate the number of moles as follows.

              No. of moles = \frac{mass}{\text{molar mass}}

                                   = \frac{55.98 g}{120.91 g/mol}

                                   = 0.462 mol

Thus, we can conclude that there are 0.462 moles present in 55.98g of CF_{2}Cl_{2}.

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The question is incomplete, here is the complete question:

Iron(III) oxide and hydrogen react to form iron and water, like this:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

At a certain temperature, a chemist finds that a 5.4 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition:

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<u>Answer:</u> The value of equilibrium constant for the given reaction is 2.8\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 3.63 g

Molar mass of hydrogen gas = 2 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

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  • <u>For water:</u>

Given mass of water = 2.13 g

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Volume of solution = 5.4 L

Putting values in above equation, we get:

\text{Molarity of water}=\frac{2.13}{18\times 5.4}\\\\\text{Molarity of water}=0.0219M

The given chemical equation follows:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of K_c for above equation follows:

K_c=\frac{[H_2O]^3}{[H_2]^3}

The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression. So, the concentration of iron and iron (III) oxide is not present in equilibrium constant expression.

Putting values in above equation, we get:

K_c=\frac{(0.0219)^3}{(0.336)^3}\\\\K_c=2.77\times 10^{-4}

Hence, the value of equilibrium constant for the given reaction is 2.8\times 10^{-4}

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