Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

use the activity series below to predict the products of each of the fo…

Question

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
f₂ > cl₂ > br₂ > i₂
ccl₄ + br₂ → ?
☑ no reaction
brcl + c
cbr₄ + cl₂
complete
feso₄ + na → ?
○ no reaction
○ na₂so₄ + fe
○ fena + so₄
done

Explanation:

For the reaction $\boldsymbol{\ce{CCl4 + Br2 ->}}$:

Step1: Check halogen activity series

The halogen activity series is $\ce{F2 > Cl2 > Br2 > I2}$. This means a halogen can displace another halogen from its compound only if it is more reactive (higher in the series).

Step2: Compare $\ce{Br2}$ and $\ce{Cl2}$ reactivity

Here, we are looking at the reaction between $\ce{CCl4}$ (containing $\ce{Cl}$) and $\ce{Br2}$. Since $\ce{Br2}$ is less reactive than $\ce{Cl2}$ (because $\ce{Cl2}$ is higher in the halogen activity series than $\ce{Br2}$), $\ce{Br2}$ cannot displace $\ce{Cl}$ from $\ce{CCl4}$. So, there is no reaction.

For the reaction $\boldsymbol{\ce{FeSO4 + Na ->}}$:

Step1: Check metal activity series

The metal activity series is $\ce{Li > K > Ba > Sr > Ca > Na > Mg > Al > Mn > Zn > Cr > Fe > Cd > Co > Ni > Sn > Pb > H > Sb > Bi > Cu > Ag > Pd > Hg > Pt > Au}$.

Step2: Compare $\ce{Na}$ and $\ce{Fe}$ reactivity

Sodium ($\ce{Na}$) is more reactive than iron ($\ce{Fe}$) because $\ce{Na}$ is higher in the metal activity series than $\ce{Fe}$. In a single - displacement reaction, a more reactive metal can displace a less reactive metal from its salt solution. So, $\ce{Na}$ will displace $\ce{Fe}$ from $\ce{FeSO4}$. The reaction will be $\ce{FeSO4 + Na -> Na2SO4 + Fe}$ (we don't need to balance it here).

Answer:

s:

  • For $\ce{CCl4 + Br2 ->}$: no reaction
  • For $\ce{FeSO4 + Na ->}$: $\ce{Na2SO4 + Fe}$ (the option is: $\boldsymbol{\ce{Na2SO4 + Fe}}$)