Answer
The balanced equation for this reaction is:
$2Na(s) + 2H_2O(l) -- \gt H_2(g) + 2NaOH(aq)$
Work Step by Step
1. Identify the chemical formula of each reactant and product:
Solid sodium: $Na(s)$
Liquid water: $H_2O(l)$
Since we added these compounds to a water solution, the hydroxide will have an $(aq)$.
Hydrogen gas: $H_2(g)$
Sodium hydroxide: $NaOH(aq)$
2. Write the equation:
$Na(s) + H_2O(l) -- \gt H_2(g) + NaOH(aq)$
3. Balance the equation:
$Na:$ Products: 1; Reactants: 1;
$O:$ Products: 1; Reactants: 1;
$H:$ Products: 3; Reactants: 2: Unbalanced:
Since we have a compound like $H_2$, that only has hydrogen in its formula, we can change its coefficient to balance the equation.
- We need to remove one hydrogen from the products, therefore, we can put a $\frac{1}{2}$ as the coefficient of $H_2$:
$Na(s) + H_2O(l) -- \gt \frac{1}{2} H_2(g) + NaOH(aq)$
Now, to remove the fraction, multiply all the coefficients by 2:
$2Na(s) + 2H_2O(l) -- \gt 2*\frac{1}{2} H_2(g) + 2NaOH(aq)$
$2Na(s) + 2H_2O(l) -- \gt H_2(g) + 2NaOH(aq)$
