solid sodium oxide and gaseous water are formed by the decomposition of solid sodium hydroxide (naoh). write a balanced chemical equation for this reaction.

Question

Accepted Answer

# Explanation:
## Step1: Write the unbalanced equation
The reactant is $NaOH$ (solid), and the products are $Na_2O$ (solid) and $H_2O$ (gaseous). So the un - balanced equation is $NaOH(s)
ightarrow Na_2O(s)+H_2O(g)$
## Step2: Balance the sodium atoms
On the left - hand side (LHS), there is 1 $Na$ atom in $NaOH$, and on the right - hand side (RHS), there are 2 $Na$ atoms in $Na_2O$. So we put a coefficient of 2 in front of $NaOH$. The equation becomes $2NaOH(s)
ightarrow Na_2O(s)+H_2O(g)$
## Step3: Balance the hydrogen and oxygen atoms
Now, for the LHS: In $2NaOH$, the number of $H$ atoms is 2 ($2\times1$ from $NaOH$) and the number of $O$ atoms is $2 + 2=4$ ($2$ from $Na - O$ and $2$ from $O - H$). On the RHS: In $Na_2O$ there is 1 $O$ atom and in $H_2O$ there are 2 $H$ atoms and 1 $O$ atom. The total $H$ atoms on RHS is 2 (from $H_2O$) and total $O$ atoms on RHS is $1 + 1=2$. The equation $2NaOH(s)
ightarrow Na_2O(s)+H_2O(g)$ is balanced for $Na$ ($2$ on both sides), $H$ ($2$ on both sides) and $O$ ($2$ on both sides)

# Answer:
$2NaOH(s)
ightarrow Na_2O(s)+H_2O(g)$

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QUESTION IMAGE

Question

Explanation:

Step1: Write the unbalanced equation

Step2: Balance the sodium atoms

Step3: Balance the hydrogen and oxygen atoms

Answer: