How do you balance the following:
KI->K + I2
Na + H2O -> H2 + NaOH
Ch4 + O2 -> CO2 + H2O
These simpler equations can be balanced by trial and error. For example, here is how you go about it.
Na + H2O ==> H2 + NaOH.
I look at the equation and i see Na is ok, O is ok, but H has 2 atoms on the left and 3 on the right. Whatever I do to balanced H will mess up Na and O but so be it.
Since I need more H on the left, and 1 doesn't work as a coefficient, we'll try 2.
Na + 2H2O ==> H2 + NaOH
Now I have 4 H on the left and I can make 4 H on the right by placing a 2 in front of NaOH; like so.
Na + 2H2O ==> H2 + 2NaOH.
Now, I still have 4 H on left and 4 on the right. I also have 2 O on the left and 2 on the right. I see I have 2 Na on the right and I can balance Na by placing a 2 in front of Na on the left like so.
2Na + 2H2O ==> H2 + 2NaOH.
Now we check it to make sure we are right.
2Na on left; 2 on the right.
4 H on the left; 4 on the right.
2 O on the left; 2 on the right.
OK. Good. It balances.
I'll leave the others for you.
To balance chemical equations, you need to ensure that the number of atoms of each element on both sides of the equation is equal. Here's how you can balance each of the given equations:
1. KI -> K + I2:
Let's count the number of iodine atoms on both sides. Initially, you have one iodine atom on the left-hand side (LHS), but on the right-hand side (RHS), you have two iodine atoms because I2 is not a single atom, but a diatomic molecule. To balance it, you can place a coefficient of 2 in front of KI:
2KI -> K + I2
Now, both sides have two iodine atoms.
2. Na + H2O -> H2 + NaOH:
First, let's balance the hydrogen atoms. On the LHS, we have 2 hydrogen atoms, while on the RHS, we only have two hydrogen atoms in H2. Therefore, the hydrogen atoms are already balanced.
Now, let's balance the sodium and oxygen atoms. On the LHS, we have one sodium atom, but on the RHS, we have one sodium atom in NaOH. Therefore, the sodium atoms are also balanced.
On the LHS, we have one oxygen atom from H2O, while on the RHS, we have one oxygen atom in NaOH and one oxygen atom in H2O, totaling two oxygen atoms. To balance it, you can place a coefficient of 2 in front of NaOH:
2Na + 2H2O -> H2 + 2NaOH
Now, the equation is balanced with all the atoms on both sides being equal.
3. CH4 + O2 -> CO2 + H2O:
Start by balancing the carbon atoms. On the LHS, we have one carbon atom, while on the RHS, we have one carbon atom in CO2. Therefore, the carbon atoms are already balanced.
Next, let's balance the hydrogen atoms. On the LHS, we have four hydrogen atoms from CH4, and on the RHS, we have two hydrogen atoms in H2O. To balance it, you can place a coefficient of 2 in front of H2O:
CH4 + O2 -> CO2 + 2H2O
Now, let's balance the oxygen atoms. On the LHS, we have two oxygen atoms from O2, while on the RHS, we have two oxygen atoms in CO2 and four oxygen atoms in 2H2O, totaling six oxygen atoms. To balance it further, you can place a coefficient of 3/2 (or 1.5) in front of O2:
CH4 + (3/2)O2 -> CO2 + 2H2O
Alternatively, you can multiply the entire equation by 2 to eliminate fractions:
2CH4 + 3O2 -> 2CO2 + 4H2O
Now, the equation is balanced with all the atoms on both sides being equal.