Question

A cobalt compound containing ammonia is analysed for determining its formula, by reacting it against a acid as follows:
$Co{\left({NH}_3\right)}_x{Cl}_2\left(aq\right)+HCl\to {NH}_4^{\oplus }\left(aq\right)+{CO}^{2+}\left(aq\right)+{Cl}^{-}\left(aq\right)$
A $3.96gm$ of this cobalt compound exactly required $40ml$ of $2M$ $HCl$ for complete reaction.
(i) Determine formula of cobalt compound
(ii) Calculate the ratio of number of moles of cation to anion produced in given reaction.

Answer 1

$Co\left(N{H}_3{)}_{x}C{l}_2\right(aq)+xHCl\rightleftharpoons xN{H}_4^{\oplus }(aq)+C{O}^{2+}(aq)+C{l}^{-}(aq)$

Here, 1 mole of Cobalt compound requires x moles of $HCl$.

Given moles of $HCl=40\times 2\times {10}^{-3}=0.08$

$\therefore \frac{3.96}{129.9+17x}$ requires $\frac{3.96x}{129.9+17x}$ moles of $HCl$.

Hence, $\frac{3.96x}{129.9+17x}=0.08$

$x(3.96-1.36)=10.392$

$\Rightarrow x\approx 4$

(i) Hence, formula of this compound is $Co(N{H}_3{)}_{4}C{l}_2$.

(ii) Moles of cation formed$=x+1=5$

Moles of anion produced$=2+x=6$.

$\therefore$ Required ratio$=\frac{5}{6}=5:6$