Iron (II) sulfate is oxidized - it means iron changes from +2 to +3
If iron is oxidized then dichromate is reduced from Cr₂O₇ ^2- in which its oxidation number is +6 to lower oxidation number.
a and b clearly out because chromium is not undergoing any change, C is also out because Fe has same +2 oxidation number.
d and e can be the option now we need some more information about Fe₂(SO₄)₃ compound. If you check the solubility list rule no 3 - all sulfates are soluble except with 6 metals -CaSO₄,BaSO₄ ,SrSO₄ Ag₂SO₄,PbSO₄ and HgSO₄.
Fe₂(SO₄)₃ is a strong electrolyte so completely soluble. So the compound can not come out as solid and it should be in solution as Fe³⁺ and SO₄^2-
Based on this understanding e is the right answer.

I don't know how to approach this question

thank you

Quote from: Aditi on April 29, 2024, 01:00:53 AMHow many grams of precipitate will form?

Solid formed is AgI.
Get the moles of AgI from BCA table and convert that it into grams.
We know moles = mass / molar mass
Rearrange the above equation
mass = moles * molar mass

Quote from: Aditi on April 29, 2024, 12:58:32 AMneed the limiting reactant as well. thank you

Limiting reactant is the one which gets completely used up or the one which gives minimum amount of the product.
Check BCA and check finally which reactant is used up completely.

Quote from: Aditi on April 29, 2024, 12:54:09 AM15.0 ml of 0.250 M potassium iodide is mixed with 40 ml of  0.170 M silver chromate.
 
set up the BCA table.
Calculate the concentrations of all ions after the reaction

15.0 ml of 0.250 M potassium iodide 
Moles of KI = moles of I- = M * V (L) = 3.75* 10^-3 mol  = moles of K⁺
40 ml of  0.170 M silver chromate.
Moles of Ag₂CrO₄  = 2moles of Ag⁺ = M * V(L) = 6.8 * 10^-3 mol = moles of CrO₄ ^2-
Moles of Ag⁺ = 6.8 * 10^-3 mol= 3.4*10^-3 mol

                    I^-(aq)    +        Ag⁺ (aq)  ->          AgI(s)
B                  3.75* 10^-3          3.4*10^-3                  0
C                  -3.4*10^-3          -3.4*10^-3              +3.4*10^-3
A                  0.35 *10^-3            0                    3.4*10^-3
Only I- and Spectator ions CrO₄ ^2- (aq) and K⁺(aq)
are  left in the solutions.
Total volume of the solution = 15.0 + 14.0 = 29.0 mL
[I^-] = 0.35 *10^-3  / 29.0
[K⁺] = 3.75* 10^-3  /  29.0

Quote from: Aditi on April 29, 2024, 12:54:09 AM15.0 ml of 0.250 M potassium iodide is mixed with 40 ml of  0.170 M silver chromate.

Please write Molecular, total, and ionic equations.
 

Balanced molecular equation is :
2KI(aq) + Ag₂CrO₄  -> K₂CrO₄ (aq) + 2AgI(s)

Now from solubility rules we know-
All K⁺ and all CrO₄^2- ions containing compounds are strong electrolytes and  AgI is not so soluble.
2K⁺(aq) + 2I^-(aq) + 2Ag⁺ (aq) + CrO₄ ^2- (aq) ->2AgI(s) +2K⁺(aq) + CrO₄ ^2- (aq)
Spectator ions are
CrO₄ ^2- (aq) and K⁺(aq)
Cancel them and net ionic equation is
2I^-(aq) + 2Ag⁺ (aq) -> 2AgI(s)

How many grams of precipitate will form?

need the limiting reactant as well. thank you

15.0 ml of 0.250 M potassium iodide is mixed with 40 ml of  0.170 M silver chromate.

Please write Molecular, total, and ionic equations.

set up the BCA table.
Calculate the concentrations of all ions after the reaction