Answer
$H_2SO_4(aq) + 2NaOH(aq) \longrightarrow 2H_2O(l) + Na_2SO_4$
Work Step by Step
1. Identify the type of Acid-Base reaction:
$NaOH$ has the $OH^-$ ion, and it is reacting with an acid ($H_2SO_4$). Therefore, this is a reaction between an acid and an hydroxide.
$Acid+Hydroxide \longrightarrow H_2O(l)+Salt$
2. Find the identity of the salt.
The $H_2SO_4$ will donate its H+ ions (since it is an acid). The "$SO_4^{2-}$" will remain in the solution.
The sodium hydroxide ($NaOH$) will donate a $OH^-$ ion, leaving the sodium ion ($Na^+$) alone.
Therefore, the produced salt is a mixture of $Na^+$ and $SO_4^{2-}$. To balance the charge, there should be 2 sodium ions in the salt: $Na_2SO_4$
3. Follow the pattern:
$H_2SO_4(aq) + NaOH(aq) \longrightarrow H_2O(l) + Na_2SO_4$
**Notice, the equation isn't balanced, there are 2 sodium atoms on the products side, and only 1 on the reactants side. To fix that, we shall put a 2 as the coefficient of $NaOH$:
$H_2SO_4(aq) + 2NaOH(aq) \longrightarrow H_2O(l) + Na_2SO_4$
** The equation isn't balanced yet, there is a total of 4 hydrogen on the reactants side, and only 2 on the products side. To fix that, we shall put a 2 as the coefficient of H2O:
$H_2SO_4(aq) + 2NaOH(aq) \longrightarrow 2H_2O(l) + Na_2SO_4$
Now, the equation is balanced.
