Question

A chemical reaction $\left( A \right)\left( {black} \right) + dilHCl{\left( B \right)_{\left( l \right)}}\xrightarrow{\Delta }{\left( C \right)_{\left( g \right)}}$. Gas$\left( C \right)$ turns lead acetate paper black. $\left( B \right)$ gives orange ppt $\left( D \right)$ soluble in excess of $KI$ forming $E$
State whether it is true or false. $\left( A \right)$ is $HgS$ .
a.True
b.False

Answer 1

We will first assume $HgS$ as $A$ in the reaction and see what we get as products.Then after we get products we will see the reaction of lead acetate paper with the gas

Complete step by step answer:
The reaction is given as follows: $\left( A \right)\left( {black} \right) + dilHCl{\left( B \right)_{\left( l \right)}}\xrightarrow{\Delta }{\left( C \right)_{\left( g \right)}}$
In this reaction, reactant $A$ which is black in colour reacts with reactant $\left( B \right)$ that is dilute hydrochloric acid to give gas $C$ which turns lead acetate paper black.
Reactant $\left( B \right)$ gives orange precipitate as the product $D$ .
When this product $D$ reacts with excess $KI$ it forms the product $E$ .
Now we will see what happens if we consider Reactant $A$ as $HgS$ .
When mercury sulphide reacts with dilute hydrochloric acid it reacts to give hydrogen sulphide and mercuric chloride.
$HgS + 2HCl \to HgC{l_2} + 2{H_2}S$
So as assumed reactant $A = HgS$ and the gas $C$ that turned lead acetate paper black is ${H_2}S$ .
When this mercuric chloride reacts $KI$ it forms mercuric iodide and gives orange color precipitate.
$HgC{l_2} + KI \to Hg{I_2}$
So, here Product $D = Hg{I_2}$
When further excess of $KI$ is added it forms mercury potassium iodide.
The reaction is given as follows:
$Hg{I_2} + K{I_{excess}} \to {K_2}Hg{I_4}$
Here, Product $E = {K_2}Hg{I_4}$
The only compound that turns lead acetate paper black is $H2S$ (hydrogen sulphide) which is a gas.
The reaction of lead acetate with hydrogen sulphide is as follows:
$Pb{\left( {C{H_3}COO} \right)_2} + {H_2}S \to PbS + C{H_3}COOH$
In this reaction lead acetate when it reacts with hydrogen sulphide it forms lead sulphide and acetic acid.
So from the above reaction we came to know what are the reactants and products.
Hence, from all this we get to know that product $A$ is $HgS$ only.

Therefore, the correct answer is a) true.

Note: Use of excess potassium iodide is used in order to solubilize free iodine ions present in the solution. Lead acetate is colorless and Hydrogen sulphide gas turns lead acetate paper black because it produces lead sulphide which is a black colour precipitate because of which it turns paper black.