What mass of potassium dichromate is required to oxidize an #100*mL# volume of #Fe^(2+)# solution whose concentration is #0.10*mol*L^-1#?

Question

Noah Aldrich · Accepted Answer

#"Approx. 1/2 a gram of potassium dichromate............"#

An oxidation equation is required:

#Fe^(2+) rarr Fe^(3+) + e^-#

Additionally, a reduction formula

#Cr_2O_7^(2-) +14H^(+) + 6e^(-) rarr 2Cr^(3+) + 7H_2O(l)#

As a whole:

#Cr_2O_7^(2-) + 6Fe^(2+) + 14H^(+)rarr 2Cr^(3+) + 6Fe^(3+) +7H_2O(l)#

#"Moles of"# #Fe^(2+):#

#100xx10^-3*Lxx0.10*mol*L^-1=0.010*mol#

And thus we need #(0.010*mol)/6xx294.18*g*mol^-1=0.490*g# #K_2Cr_2O_7#.

Jose Ayala · Answer

undefined To determine the mass of potassium dichromate required to oxidize the given volume of Fe^(2+) solution, we need to first balance the redox equation between Fe^(2+) and dichromate ions (Cr2O7^2-). The balanced equation is:

6 Fe^(2+) + Cr2O7^2- + 14 H^+ -> 6 Fe^(3+) + 2 Cr^(3+) + 7 H2O

From the balanced equation, we can see that 1 mole of dichromate ion (Cr2O7^2-) oxidizes 6 moles of Fe^(2+).

Given that the concentration of Fe^(2+) is 0.10 mol/L and the volume is 100 mL (which is 0.1 L), the number of moles of Fe^(2+) present is:

0.10 mol/L * 0.1 L = 0.01 moles

Since 1 mole of dichromate ion oxidizes 6 moles of Fe^(2+), the number of moles of dichromate required is:

0.01 moles * (1/6) = 0.00167 moles

Now, we can calculate the mass of potassium dichromate using its molar mass, which is 294.185 g/mol. The mass (m) is given by:

m = number of moles * molar mass

m = 0.00167 moles * 294.185 g/mol

m ≈ 0.491 g

Therefore, approximately 0.491 grams of potassium dichromate is required to oxidize the given volume of Fe^(2+) solution.