How do i balance this equation out?
Na2S2O2 + I2 ---> NaI+ Na2S4O6
thank you soo much! =D
You made a typo. You meant Na2S2O3, I'm sure. So I changed it below and balanced it. By the way, you need to learn to do this on your own, especially since this one is so simple. The four S atoms on the left have an oxidation number of +8 and they change to +10 on the right. The I2 on the left is zero and on the right is -2 for both of the I atoms.
2Na2S2O3 + I2 ---> 2NaI+ Na2S4O6
To balance the equation:
1. Count the number of atoms of each element on both sides of the equation. Atoms that appear in more than one compound should be counted separately.
On the left side:
- Sodium (Na): 2 atoms
- Sulfur (S): 2 atoms
- Oxygen (O): 2 atoms
On the right side:
- Sodium (Na): 3 atoms
- Sulfur (S): 4 atoms
- Oxygen (O): 8 atoms
- Iodine (I): 1 atom
2. Start balancing the equation with the elements that appear in the fewest compounds. In this case, it's iodine, which appears only once on the right side.
Add a coefficient of 2 in front of NaI on the right side to balance the iodine:
Na2S2O2 + I2 ---> 2NaI + Na2S4O6
3. Next, balance the sodium atoms. On the left side, there are two sodium atoms, while on the right side, there are three. Add a coefficient of 3 in front of NaI on the right side to balance the sodium:
Na2S2O2 + I2 ---> 3NaI + Na2S4O6
4. Finally, balance the sulfur and oxygen atoms. On the right side, there are four sulfur atoms, while on the left side, there are only two. Multiply the coefficient in front of Na2S2O2 on the left side by 2 to balance the sulfur:
2Na2S2O2 + I2 ---> 3NaI + Na2S4O6
Now, the equation is balanced with:
2Na2S2O2 + I2 ---> 3NaI + Na2S4O6.