Question

2na (s) + cl₂ (g) → 2nacl (s)
2na₂o (s) → 4na (s) + o₂ (g)
in the final chemical equation, nacl and o₂ are the products that are formed through the reaction between na₂o and cl₂. before you can add these intermediate chemical equations, you need to alter them by
multiplying the second equation by 2.
multiplying the first equation by 2.
multiplying the second equation by (1/4).
multiplying the first equation by (1/2).

Explanation:

Step1: Analyze the target products

We want to get $NaCl$ and $O_2$ as products from the reaction of $Na_2O$ and $Cl_2$. The first equation gives $NaCl$ from $Na$ and $Cl_2$ and the second gives $Na$ and $O_2$ from $Na_2O$.

Step2: Match the amounts of $Na$

In the first - equation, 2 moles of $Na$ react to form 2 moles of $NaCl$. In the second equation, 4 moles of $Na$ are produced from 2 moles of $Na_2O$. To make the amount of $Na$ consistent between the two equations (so that we can cancel out $Na$ when adding the equations), we need to multiply the first equation by 2. Then we will have 4 moles of $Na$ in the first equation coming from the reaction of $2Cl_2$ with 4 moles of $Na$, and 4 moles of $Na$ in the second equation coming from 2 moles of $Na_2O$.

Answer:

multiplying the first equation by 2.