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3 years ago

Fe(NO3)2 not sure how to get the oxidation numbers of all elements

1 answer:

4 0

You must remember that oxidation number of hydrogen in acids is always +1, oxidation number of oxygen in oxides & acids is always -2... metals has always oxidation number on plus!

group NO3 comes from HNO3...and oxidation number of whole acid group is always on minus and equal to the amount of hydrogen atoms in this acid... so oxidation number of NO3 = -1

we have 2 NO3 groups so 2*(-1) = -2 and that is the reason why oxidation number of Fe in this formula must be +2... because sum of all elements always gives 0!

Now we could count of oxidation number for nitrogen... we write HNO3 and start counting from right to left:
3*(-2) from oxygens + 1 from hydrogen = -5
so nitrogen must have +5 oxidation number... because sum all in formula must be 0.

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<u>Answer:</u> The moles of $CO_2$ added to the system is 7.13 moles

<u>Explanation:</u>

We are given:

Moles of $CO_2$ at equilibrium = 1.00 moles

Moles of $H_2$ at equilibrium = 1.00 moles

Moles of $H_2O$ at equilibrium = 2.40 moles

Moles of $CO$ at equilibrium = 2.40 moles

Volume of the container = 4.00 L

Concentration is written as:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The given chemical equation follows:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{2.40}{4.00})\times (\frac{2.40}{4.00})}{(\frac{1.00}{4.00})\times (\frac{1.00}{4.00})}\\\\K_c=5.76$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of CO = 0.791 mol/L

Volume of solution = 4.00 L

Putting values in above equation, we get:

$0.791M=\frac{\text{Moles of CO}}{4.00L}\\\\\text{Moles of CO}=(0.791mol/\times 4.00L)=3.164mol$

Extra moles of CO = (3.164 - 2.40) = 0.764 moles

Let the moles of $CO_2$ needed be 'x' moles.

Now, equilibrium gets re-established:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

Initial: 1.00 1.00 2.40 2.40

At eqllm: (0.236+x) 0.236 3.164 3.164

Again, putting the values in the expression of $K_c$ , we get:

$5.76=\frac{(\frac{3.164}{4.00})\times (\frac{3.164}{4.00})}{(\frac{0.236+x}{4.00})\times (\frac{0.236}{4.00})}\\\\5.76=\frac{10.011}{0.056+0.236x}\\\\x=7.13$

Hence, the moles of $CO_2$ added to the system is 7.13 moles

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3 years ago