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Zanzabum
3 years ago
15

What is the balanced chemical equations for fe + o2 → iron(౹౹) oxide?

Chemistry
1 answer:
pishuonlain [190]3 years ago
3 0
The idea behind balancing chemical equations is that the number of atoms an element has on the reactants' side must be equal to the number of atoms it has on the products' side.

These atoms will become a part of different compounds once the reaction is completed, but they must always be in equal numbers on both sides.

So, look at iron first. One atom reacts, but two are produced - notice the 2 subscript iron has in Fe2O3. This means you must double the number of atoms on the reactants' side to reach an equality.

2Fe(s)+O2(g)→Fe2O3(s)

Now look at oxygen. Two atoms react, but three are produced. The trick here is to find a common multiple that will make the number of atoms equal on both sides.

The easiest way to do this is to multiply the atoms that react by 3, which will give you 6 oxygen atoms that react, and the atoms that are produced by 2 - this will get you 6 oxygen atoms produced.

2Fe(s)+3O2(g)→2Fe2O3(s)

However, notice that the iron atoms are unbalanced again. You have 2 that react, but 4 that are produced → multiply the atoms that react by 2 again, which will give you

4Fe(s)+3O2(g)→2Fe2O3(s)

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3 years ago
Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s
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This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, 2.67\times 10^2mol/L

Explanation :

As we know that,

C_{O_2}=k_H\times p_{O_2}

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C_{O_2} = molar solubility of O_2 = ?

p_{O_2} = partial pressure of O_2 = 0.2 atm  = 1.97×10⁻⁶ Pa

k_H = Henry's law constant  = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)

C_{O_2}=8.55\times 10^3g/L

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = \frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L

Therefore, the molar concentration of oxygen is, 2.67\times 10^2mol/L

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