The number of ions present in $2.0 \ L$ of a solution of $0.8 \ M \ K_4[Fe(CN)_6]$ is

When potassium iodide is added to an aqueous solution of potassium ferricyanide,a reversible reaction is observed in which a complex $P$ is formed. In a strong acidic medium,the equilibrium shifts completely towards $P$. Addition of zinc chloride to $P$ in a slightly acidic medium results in a sparingly soluble complex $Q$.
$(1)$ The number of moles of potassium iodide required to produce two moles of $P$ is. . .
$(2)$ The number of zinc ions present in the molecular formula of $Q$ is. . . .
Give the answer for $(1)$ and $(2)$.

The reaction of $K_3[Fe(CN)_6]$ with freshly prepared $FeSO_4$ solution produces a dark blue precipitate called Turnbull's blue. The reaction of $K_4[Fe(CN)_6]$ with $FeSO_4$ solution in the complete absence of air produces a white precipitate $X$,which turns blue in air. Mixing the $FeSO_4$ solution with $NaNO_3$,followed by a slow addition of concentrated $H_2SO_4$ through the side of the test tube,produces a brown ring.
Precipitate $X$ is:
$(A)$ $Fe_4[Fe(CN)_6]_3$
$(B)$ $Fe[Fe(CN)_6]$
$(C)$ $K_2Fe[Fe(CN)_6]$
$(D)$ $KFe[Fe(CN)_6]$
Among the following,the brown ring is due to the formation of:
$(A)$ $[Fe(NO)_2(SO_4)_2]^{2-}$
$(B)$ $[Fe(NO)_2(H_2O)_4]^{3+}$
$(C)$ $[Fe(NO)_4(SO_4)_2]$
$(D)$ $[Fe(H_2O)_5(NO)]SO_4$

$A$ mixture $X$ of $0.02 \, mol$ of $[Co(NH_3)_5SO_4]Br$ and $0.02 \, mol$ of $[Co(NH_3)_5Br]SO_4$ was prepared in $2 \, L$ of solution:
$1 \, L$ of mixture $X +$ excess of $AgNO_3 \to Y$
$1 \, L$ of mixture $X +$ excess of $BaCl_2 \to Z$
Number of moles of $Y$ and $Z$ respectively are

For the reaction $K_4[Fe(CN)_6] \xrightarrow{\text{oxidation}} Fe^{3+} + CO_2 + NO_3^-$,which of the following statements is incorrect?

For the given aqueous reaction,which of the statement$(s)$ is (are) true?
$2KI + 2K_3[Fe(CN)_6] \xrightarrow{H_2SO_4} I_2 + 2K_4[Fe(CN)_6]$
$(A)$ The first reaction is a redox reaction.
$(B)$ White precipitate is $K_2Zn_3[Fe(CN)_6]_2$.
$(C)$ Addition of filtrate to starch solution gives blue colour.
$(D)$ White precipitate is soluble in $NaOH$ solution.