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Veronika [31]
3 years ago
10

A sealed 1.0L flask is filled with 0.500 mols of I_2 and 0.500 mols of Br_2. When the container achieves equilibrium the equilib

rium constant is 1.10x10^{-2}. What is the equilibrium concentration of the product, IBr?
Chemistry
1 answer:
DochEvi [55]3 years ago
7 0

Answer:

[IBr] = 0.049 M.

Explanation:

Hello there!

In this case, according to the balanced chemical reaction:

I_2+Br_2\rightarrow 2IBr

It is possible to set up the following equilibrium expression:

K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110

Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of x (reaction extent) would be:

0.0110=\frac{(2x)^2}{(0.50-x)^2}

Which can be solved for x to obtain two possible results:

x_1=-0.0277M\\\\x_2=0.0245M

Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:

[IBr]=2x=2*0.0249M=0.049M

Regards!

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1. True or False: Fungi are reproduced through the spread of spores.<br> O True<br> O False
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Read 2 more answers
If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of
Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.1\ g}{186.04\ g/mol}

Moles\ of\ ferrocene= 0.0005375\ mol

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

\rho=\frac{Mass}{Volume}

or,  

Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.044\ g}{78.49\ g/mol}

Moles\ of\ acetyl\ chloride= 0.0005606\ mol

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms  0.0005375 mole of monoacetylferrocene

Moles of product formed =  0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %</u>

5 0
3 years ago
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