Question

Solid lead(ii) sulfide reacts with aqueous hydrochloric acid to form solid lead(ii) chloride and dihydrogen sulfide gas. express your answer as a chemical equation. identify all of the phases in your answer.

Answer 1

Answer: The chemical equation is;

$PbS(s)+2HCl(aq)\rightarrow PbCl_2(s)+H_2S(g)$

Explanation:

The described reaction between solid lead sulfide and aqueous hydrochloric acid can represented in the in the form of chemical equation as given below:

$PbS(s)+2HCl(aq)\rightarrow PbCl_2(s)+H_2S(g)$

1 mol of lead(ii) sulfide reacts with hydrochloric acid to give 1 mole of lead chloride and hydrogen sulfide gas.

Answer 2

The reaction expressed as a chemical equation is as follows;

• PbS(s) + HCl(aq) ====> PbCl2(s) + H2S(g).

According to the question;

The reactants are;

• Solid lead(ii) sulfide and aqueous hydrochloric acid

The products are;

• solid lead(ii) chloride and dihydrogen sulfide gas

In essence, the reaction is thus;

• PbS(s) + HCl(aq) ====> PbCl2(s) + H2S(g).

The phases in the chemical equation are as follows;

• PbS = Solid phase
• PbS = Solid phaseHCl = Aqueous phase
• PbS = Solid phaseHCl = Aqueous phasePbCl2 = Solid phase
• PbS = Solid phaseHCl = Aqueous phasePbCl2 = Solid phaseH2S = Gaseous phase.

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