chemistry

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Can some one please balance this equation for me
Fe + O2---> Fe2O3.

Fe + O2 ==> Fe2O3.
Here is the way to balance these equations by inspection, meaning, by trial and error.
Rule #1. You may not change any subscripts. Equations are balanced by changing ONLY the coefficients.
Look on the right side and I see 2 Fe atoms. So we move to the left and place a 2 in front of the Fe.
2Fe + O2 ==> Fe2O3

Now look at the 3 O on the right. That is what we must have on the left. Two ways to approach this. I can balance it by placing 3/2 in front of the O2 like this.
2Fe + 3/2 O2 ==> Fe2O3. And the equation balances. Let's check it.
We have 2 Fe atoms on the left and right.
We have 3 O (3/2 * 2 = 3) atoms on the left and right.
Most of us like to see equations in whole numbers, not fractions, so we now multiply the entire equation by 2 to get rid of the 1/2.
4Fe + 3O2 ==> 2Fe2O3.

The second way: Starting with
2Fe + O2 ==> Fe2O3.

Look at O on the right and there are 3 atoms with two O atoms on the left. We know a whole numbrer coefficient won't work, so we trial and error. Knowing that 6 is a multiple of both 2 and 3, we just multiply Fe2O3 by 2 in order to have 6 O atoms on the right. And on the left we have a coefficient of 3 for O2.
2Fe + 3O2 ==> 2Fe2O3.
That gives us 6 O on the left and 6 on the right. Of course, it messed up the Fe but we can fix that easily. Since we now have 4 Fe on the right, we change the 2 for 2Fe on the left to 4Fe and the final equation is
4Fe + 3O2 ==> 2Fe2O3.

If your algebra is up to speed the balancing can also be done this way:

Fe + xO2 ==> yFe2O3

balancing Fe atoms
1=2y------------------(1

balancing O atoms
2x=3y------------------(2

for from equation (1
y=1/2

and substituting this in equation (2

2x=3(1/2)

so x=3/4

Replacing the symbols in the original

Fe + 3/4O2 ==> 1/2Fe2O3

but we are not allowed fractions so multiply through by 4

4Fe + 3O2 ==> 2Fe2O3

  • chemistry -

    Ag+
    N2==>Ag3N

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