Question

For the following reaction between Mohr's salt (iron as FeSO_4(NH_4)_2SO_4 6H_2O) and potassium dichromate (dichromate as K_2Cr_2O_7), determine the volume (in milliliters) of a 0.230 M solution of Mohr's salt that is needed to fully react with 0.0500 L of 0.230 M potassium dichromate. (The reaction is shown in its ionic form in the presence of a strong acid.)
Cr_2O_7^2- + 6Fe^2+ + 14H^+ rightarrow 2Cr^3+ + 6Fe^3+ + 7H_2O
Mohr's salt volume = mL
For the same reaction, what volume (in milliliters) of 0.230 M potassium dichromate is required to fully react with 0.0500 L of a 0.230 M solution of Mohr's salt
potassium dichromate volume = mL

Answer 1

Answer :

Part 1 : The volume (in milliliters) of a 0.230 M solution of Mohr's salt is 8.33 mL.

Part 2 : The volume (in milliliters) of a 0.230 M solution of potassium dichromate required is 300 mL.

Explanation :

Part 1 :

Using neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of $Cr_2O_7^{2-}$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of $Fe^{2+]$

We are given:

$n_1=1\\M_1=0.230M\\V_1=0.0500L\\n_2=6\\M_2=0.230M\\V_2=?$

Putting values in above equation, we get:

$1\times 0.230M\times 0.0500L=6\times 0.230M\times V_2\\\\V_2=8.33\times 10^{-3}L=8.33mL$

Conversion used : (1 L = 1000 mL)

Thus, the volume (in milliliters) of a 0.230 M solution of Mohr's salt is 8.33 mL.

Part 2 :

Using neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of $Cr_2O_7^{2-}$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of $Fe^{2+]$

We are given:

$n_1=1\\M_1=0.230M\\V_1=?\\n_2=6\\M_2=0.230M\\V_2=0.0500L$

Putting values in above equation, we get:

$1\times 0.230M\times V_1=6\times 0.230M\times 0.0500L\\\\V_1=0.3L=300mL$

Conversion used : (1 L = 1000 mL)

Thus, the volume (in milliliters) of a 0.230 M solution of potassium dichromate required is 300 mL.