Question

fill in the missing words. if the following reaction is balanced correctly $na_2o + h_2o
ightarrow naoh$, the coefficient in front of naoh is , and the coefficient in front of $h_2o$ is .

Explanation:

Step1: Balance sodium atoms

On the left - hand side, there are 2 sodium atoms in $Na_2O$. To balance sodium atoms in $NaOH$ on the right - hand side, we need 2 moles of $NaOH$. So the coefficient of $NaOH$ is 2. The unbalanced equation becomes $Na_2O + H_2O
ightarrow 2NaOH$.

Step2: Balance hydrogen and oxygen atoms

Now, on the right - hand side, there are 2 hydrogen atoms and 2 oxygen atoms in 2 moles of $NaOH$. On the left - hand side, in $Na_2O$ there is 1 oxygen atom and in $H_2O$ there are 2 hydrogen atoms and 1 oxygen atom. The equation $Na_2O + H_2O
ightarrow 2NaOH$ is balanced as it is. So the coefficient of $H_2O$ is 1.

Answer:

The coefficient in front of $NaOH$ is 2, and the coefficient in front of $H_2O$ is 1.