na + cl₂→ nacl
total # before (reactants) elements total # after (products)
na
cl

Question

Accepted Answer

# Explanation:
## Step1: Balance chlorine atoms
In $Cl_2$, there are 2 chlorine atoms. In $NaCl$, there is 1 chlorine atom. To balance chlorine, put 2 in front of $NaCl$.
$$\text{Na}+\text{Cl}_2
ightarrow 2\text{NaCl}$$
## Step2: Balance sodium atoms
After balancing chlorine, there are 2 sodium atoms on the product side ($2\text{NaCl}$). So put 2 in front of $Na$ on the reactant side.
$$2\text{Na}+\text{Cl}_2
ightarrow 2\text{NaCl}$$

For the table:
- For $Na$:
    - Reactants: Before balancing (if we consider initial count as per un - balanced equation $Na + Cl_2
ightarrow NaCl$), assume coefficient of $Na$ is $x$, coefficient of $Cl_2$ is $y$, coefficient of $NaCl$ is $z$. Initially $x = 1$, $y = 1$, $z = 1$. After balancing $x = 2$, $y = 1$, $z = 2$. So total $\#$ before (reactants) for $Na$ is $2$ (after balancing the whole equation, we look at the coefficient of $Na$ in the balanced equation $2Na+Cl_2
ightarrow 2NaCl$), total $\#$ after (products) for $Na$ is $2$ (from $2NaCl$).
 - For $Cl$:
    - Reactants: In $Cl_2$, the number of $Cl$ atoms is $2$ (since $Cl_2$ has 2 $Cl$ atoms and its coefficient is $1$ in the balanced equation $2Na + Cl_2
ightarrow 2NaCl$). Products: In $2NaCl$, the number of $Cl$ atoms is $2$ (each $NaCl$ has 1 $Cl$ atom and coefficient of $NaCl$ is $2$).

# Answer:
The balanced equation is $2\text{Na}+\text{Cl}_2
ightarrow 2\text{NaCl}$.
- For $Na$ in the table: Total $\#$ Before (reactants) $=2$, Total $\#$ After (Products) $=2$.
- For $Cl$ in the table: Total $\#$ Before (reactants) $=2$, Total $\#$ After (Products) $=2$.

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QUESTION IMAGE

Question

Explanation:

Step1: Balance chlorine atoms

Step2: Balance sodium atoms

Answer: