balance
FeS2 + O2 < Fe2O3 + So2
I assume the arrow is pointing the wrong way.
FeS2 + O2 ==> Fe2O3 + SO2
balanced is
4FeS2 + 11O2 ==> 2Fe2O3 + 8SO2
check it.
4Fe on left and 4 on right
8S on left and 8 on right.
22 O on left and (2*3=6 + 8*2=16 and 16 + 6 = 22) O on the right.
The given chemical equation represents a reaction between iron(II) sulfide (FeS2) and oxygen gas (O2) to form iron(III) oxide (Fe2O3) and sulfur dioxide (SO2). To balance this equation, you need to ensure that the number of atoms on both sides of the equation is equal.
Here's how you can balance the equation:
1. Start by counting the number of atoms present for each element on both sides of the equation. Let's start with iron (Fe).
On the left side: 1 Fe
On the right side: 2 Fe
Since there are double the number of Fe atoms on the right side, add a coefficient of 2 in front of FeS2 to balance the iron atoms.
2 FeS2 + O2 < Fe2O3 + SO2
2. Next, balance the sulfur (S) atoms.
On the left side: 2 S
On the right side: 1 S
To achieve an equal number of sulfur atoms, add a coefficient of 2 in front of SO2.
2 FeS2 + O2 < Fe2O3 + 2 SO2
3. Now, proceed to balance the oxygen (O) atoms.
On the left side: 2 O (from FeS2) + 2 O (from O2) = 4 O
On the right side: 3 O (from Fe2O3) + 4 O (from 2 SO2) = 7 O
To balance the oxygen, add a coefficient of 7/2 or 3.5 in front of O2. However, since we want to keep the coefficients as whole numbers, multiply the entire equation by 2 to eliminate the fractional coefficient.
4 FeS2 + 7 O2 < 2 Fe2O3 + 4 SO2
Now, the equation is balanced with 4 FeS2 + 7 O2 reacting to produce 2 Fe2O3 + 4 SO2.