Question

consider the following intermediate chemical equations.
2na(s)+cl2(g)→2nacl(s)
2na(s)+o2(g)→na2o2(s)
in the final chemical equation, nacl and o2 are the products that are formed through the reaction between na2o2 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 first equation by (1/2).
multiplying the second equation by (1/4).

Explanation:

Step1: Analyze the goal

We want to combine the two given intermediate chemical - equations to get the final chemical equation.

Step2: Consider stoichiometry

We need to make the amounts of intermediate species (if any) cancel out when adding the equations.

Step3: Analyze the options

If we multiply the first equation by $\frac{1}{2}$ and the second equation by $\frac{1}{4}$, we can adjust the coefficients of the products and reactants to combine the equations properly. When we multiply the first equation $2Na(s)+Cl_{2}(g)
ightarrow 2NaCl(s)$ by $\frac{1}{2}$, we get $Na(s)+\frac{1}{2}Cl_{2}(g)
ightarrow NaCl(s)$. When we multiply the second equation $2Na_{2}O_{2}(s)
ightarrow 4Na(s)+O_{2}(g)$ by $\frac{1}{4}$, we get $\frac{1}{2}Na_{2}O_{2}(s)
ightarrow Na(s)+\frac{1}{4}O_{2}(g)$. Then we can add these two new - form equations to get the overall reaction in terms of the desired products and reactants.

Answer:

multiplying the first equation by (1/2) and the second equation by (1/4)