Question

making activity series predictions
use the activity series below to predict the products of each of the following reactions. you do not need to balance the equations.
li > k > ba > sr > ca > na > mg > al > mn > zn > cr > fe > cd > co > ni > sn > pb > h > sb > bi > cu > ag > pd > hg > pt > au
$f_{2} > cl_{2} > br_{2} > i_{2}$

1. $ccl_{4}+br_{2} \

ightarrow ?$

• no reaction
• $brcl + c$
• $cbr_{4}+cl_{2}$

1. $feso_{4}+na \

ightarrow ?$

• no reaction
• $na_{2}so_{4}+fe$
• $fena + so_{4}$

Explanation:

Step1: Analyze $\text{CCl}_4 + \text{Br}_2$

From the halogen activity series: $\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2$. A more active halogen replaces a less active one in a compound. $\text{Br}_2$ is less active than $\text{Cl}_2$, so it cannot displace $\text{Cl}_2$ from $\text{CCl}_4$.

Step2: Analyze $\text{FeSO}_4 + \text{Na}$

From the metal activity series: $\text{Na} > \text{Fe}$. A more active metal replaces a less active metal in a salt. $\text{Na}$ will displace $\text{Fe}$ from $\text{FeSO}_4$, forming $\text{Na}_2\text{SO}_4$ and $\text{Fe}$.

Answer:

1. $\text{CCl}_4 + \text{Br}_2$: no reaction
2. $\text{FeSO}_4 + \text{Na}$: $\text{Na}_2\text{SO}_4 + \text{Fe}$