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Alinara [238K]
3 years ago
10

What is the total number of moles of sulfur atoms in 1 mole of Fe2(SO4)3

Chemistry
2 answers:
Komok [63]3 years ago
7 0
Since you have 3 atoms of sulfur in that molecule in on mole of it you would have 3 moles of sulfur
IgorC [24]3 years ago
4 0

<u>Answer: </u>There are 3 moles of sulfur atoms present in 1 mole of Fe_2(SO_4)_3

<u>Explanation:</u>

In the given compound Fe_2(SO_4)_3, there are 3 elements present: iron, sulfur and oxygen.

There are 2 moles of iron element, 3 moles of sulfur element and 12 moles of oxygen element present in 1 mole of Fe_2(SO_4)_3

Hence, there are 3 moles of sulfur atoms present in 1 mole of Fe_2(SO_4)_3

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<u>Answer:</u> The value of K_{eq} is 4.84\times 10^{-5}

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of ammonia = \frac{0.0280}{1.00}=0.0280M

Concentration of oxygen gas = \frac{0.0120}{1.00}=0.0120M

The given chemical equation follows:

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<u>Initial:</u>        0.0280        0.0120

<u>At eqllm:</u>    0.0280-4x   0.0120-3x   2x       6x

We are given:

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Equilibrium concentration of oxygen gas = 0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M

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The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}

Putting values in above expression, we get:

K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}

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