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Artist 52 [7]
3 years ago
9

Analysis of a sample of an oxide of irons finds that there are 87.47 g of iron and 12.53 g of oxygen present. Determine the empi

rical formula of the compound. What is iron's subscript in the empirical formula?
Chemistry
1 answer:
S_A_V [24]3 years ago
8 0

Answer:

Empirical formula:

Fe₂O

Subscript of iron in empirical formula is 2.

Explanation:

Empirical formula:

It is the simplest formula  that gives the smallest whole number ratio between the atoms of elements present in a compound.

Given data:

Mass of iron = 87.47 g

Mass of oxygen = 12.53 g

Empirical formula = ?

Solution:

Number of gram atoms of iron = 87.47 /55.845 =1.57 grams atoms

Number of gram atoms of oxygen = 12.53 / 16 = 0.78 grams atoms

Atomic ratio:

Fe                   :          O

1.57/0.78        :       0.78/0.78  

     2               :              1                    

Fe : O = 2  :  1

Empirical formula:

Fe₂O

Subscript of iron in empirical formula is 2.

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If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
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Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

<em>SO2Cl2 -----> SO2(g) + Cl2(g) </em>

<em>Kc= 2.99 x 10^-7 at 227 degrees celcius </em>

<em>If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?</em>

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 ---------> SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

<em>[Cl₂] = 2.24x10⁻⁴ M</em>

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Answer:

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