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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

need the limiting reactant as well. thank you

How many grams of precipitate will form?

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)

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, 12:58:32 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, 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

thank you