Answer
2Na +2 $H_{2}$O $\longrightarrow$ 2NaOH + $H_{2}$
Work Step by Step
Na + $H_{2}$O $\longrightarrow$ NaOH + $H_{2}$
In this chemical reaction the metal of Na enters in reaction with water, and sodium hydroxide is formed and hydrogen gas is released. To note here is that Na enters as a pure substance, and another pure substance is formed (hydrogen). So in this reaction, atoms of Na are changed into ions of Na+, while H+ is changed to gas $H_{2}$ with the charge 0.
2 H+ $\longrightarrow$ $H_{2}^{0}$
$Na^{0}$ $\longrightarrow$ $Na^{+}$
To balance the charge in the above equations, we add 2 before Na+ .
2 H+ $\longrightarrow$ $H_{2}^{0}$
2$Na^{0}$ $\longrightarrow$ 2$Na^{+}$
The same coefficient 2 we add in the chemical equation:
2Na + $H_{2}$O $\longrightarrow$ 2NaOH + $H_{2}$
Now we have to balance number of O atoms, as there is 1 on the left and 2 on the right. We add 2 before $H_{2}$O.
2Na + 2$H_{2}$O $\longrightarrow$ 2NaOH + $H_{2}$
Count the atoms of each elements and find our that we have balanced the equation.
