Question

A is a binary compound of a univalent metal. When 1.422 g of A reacts completely with 0.321 g of Sulphur in an evacuated in a sealed tube, 1.743 g of a white crystalline solid B produces, which produces a hydrated double salt C with $A{l_2}{(S{O_4})_3}$. Identify A, B and C.

Answer 1

Hint: It is given that the salt formed contains sulphate anion, that means the binary compound contains one metal cation and oxygen anion as it reacts with Sulphur to form sulphate compound.

Complete step by step answer:
Given,
Mass of compound A is 1.422 g
Mass of Sulphur is 0.321 g.
Mass of B is 1.743 g.
A is the binary compound of a univalent metal means that it contains only one metal and another atom. Here A reacts with Sulphur to form a sulphate compound, that means the binary compound contains oxygen, so the formula of the binary compound is $M{O_2}$where M is positive cation and oxygen is present as superoxide $O_2^ -$. The hydrated double salt will be ${M_2}S{O_4}$with $A{l_2}{(S{O_4})_3}$
0.321g of sulfur gives 1.743 g of metal sulphate.
$32.1g\;sulphur \Rightarrow \dfrac{{1.743}}{{0.321}} \times 32.1 = 174.3g\;metal\;sulphate$
The molecular weight of M in ${M_2}S{O_4}$ is calculated as shown below.
$\Rightarrow$2M + 32 +64 = 174.3g
$\Rightarrow$2M + 96 = 174.3g
And hence on doing the simplification,we have
$\Rightarrow$2M = 174.3g -96
$\Rightarrow$2M = 78.3
Now again on solving, we have
$\Rightarrow M = \dfrac{{78.3}}{2}$
$\Rightarrow M = 39.15$
The atomic mass 39.15 is of potassium.
From the formula of ${K_2}S{O_4}$it can be seen that Sulphur is reacted with 2 mole of $K{O_2}$.
The reaction is shown below.
$2K{O_2} + S \to {K_2}S{O_4}$
0.321g Sulphur $\Rightarrow$1.422 g metal oxide
$32.1g\;sulphur \Rightarrow \dfrac{{1.422}}{{0.321}} \times 32.1 = 142.2g\;metal\;oxide$
The molecular weight of 2$K{O_2}$ is
$\Rightarrow 2(39 + 32)$
$\Rightarrow 142$
The compound A is $K{O_2}$
The compound B is ${K_2}S{O_4}$.
The compound C is ${K_2}S{O_4}$.$A{l_2}{(S{O_4})_3}$.24${H_2}O$.

Note: Make sure to calculate the molecular weight of the metal oxide and metal sulphate to confirm the compound after knowing the metal used. Hydrated double salt contains two metal salts with molecules of water which can be removed by heating.