Answer
(a) 2N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 3N$_2$ + 4H$_2$O
(b) 2F$_2$ + 2H$_2$O ➔ 4HF + O$_2$
(c) Na$_2$O + H$_2$O ➔ 2NaOH
Work Step by Step
(a) N$_2$H$_{4}$ + N$_2$O$_4$ ➔ N$_2$ + H$_2$O
We first look at the nitrogen atoms. We have 4 on the left and 2 on the right. We add a 2 in front of the N$_2$ to balance the nitrogen.
N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 2N$_2$ + H$_2$O
We now look at the hydrogen atoms. We have 4 on the left and 2 on the right. We place a 2 in front of the H$_2$O on the left side to balance the hydrogen.
N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 2N$_2$ + 2H$_2$O
Finally, we look at the oxygen atoms. We have 4 on the left and 2 on the right. If we replace the 2 in front of the H$_2$O with a 4. The oxygen atoms are balanced.
N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 2N$_2$ + 4H$_2$O
The hydrogen are now out of balance. We have 4 on the left and 8 on the right. Let's place a 2 in front of N$_2$H$_{4}$.
2N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 2N$_2$ + 4H$_2$O
The nitrogen are now out of balance. We have 6 on the left and 4 on the right. If we change the 2 to a 3 in front of 2N$_2$, then the nitrogen will be balanced.
2N$_2$H$_{4}$ + N$_2$O$_4$ ➔ 3N$_2$ + 4H$_2$O
Now let's check the numbers of atoms in the reactants and the products.
Reactants/Products:
nitrogen: 6/6
hydrogen: 8/8
oxygen: 4/4
(b) F$_2$ + H$_2$O ➔ HF + O$_2$
We look first at the hydrogen atoms. We have 2 on the left and 1 on the right. We add 2 as a coefficient in front of the HF on the right to balance the hydrogen atoms.
F$_2$ + H$_2$O ➔ 2HF + O$_2$
We now look at the oxygen atoms. We have 1 on the left and 2 on the right. If we add a 2 in front of the H$_2$O on the left, then we balance the oxygen atoms.
F$_2$ + 2H$_2$O ➔ 2HF + O$_2$
However, now we have an imbalance of hydrogen atoms: 4 on the left and 2 on the right. Let's change the coefficient in front of HF from a 2 to a 4.
F$_2$ + 2H$_2$O ➔ 4HF + O$_2$
Finally, we look at the fluorine atoms. We have 2 on the left and 4 on the right. Let's add a 2 in front of F$_2$ to balance the fluorine atoms.
2F$_2$ + 2H$_2$O ➔ 4HF + O$_2$
Now let's check the numbers of atoms in the reactants and the products.
Reactants/Products:
fluorine: 4/4
hydrogen: 4/4
oxygen: 2/2
(c) Na$_2$O + H$_2$O ➔ NaOH
We first look at the sodium atoms. We have 2 on the left and 1 on the right. If we add a 2 in front of NaOH, we will be balanced with respect to sodium.
Na$_2$O + H$_2$O ➔ 2NaOH
We now look at the hydrogen atoms. We have 2 on the left and 2 on the right. We are balanced in terms of hydrogen.
Next, we look at the oxygen atoms. We have 2 on the left and 2 on the right. We are balanced in terms of oxygen.
Now let's check the numbers of atoms in the reactants and the products.
Reactants/Products:
sodium: 2/2
hydrogen: 2/2
oxygen: 2/2
