+3 votes
in Class 10 by kratos

Explain the action of dilute hydrochloric acid on the following with chemical equation :

(i) magnesium ribbon

(ii) sodium hydroxide

(iii) crushed egg shells

1 Answer

+5 votes
by kratos
 
Best answer

(i) Mg + 2HCI → MgCl2 + H2

Hydrogen gas is produced.

(ii) HCl + NaOH→ NaCl + H2O

Neutralisation reaction

(iii) 2HCl(aq) + CaCO3(*)→CaCl2(aq)+ H2O + CO2

Calcium chloride is formed.

Detailed Answer:

(i) Dilute hydrochloric acid reacts with magnesium to form magnesium chloride and H2 gas is liberated.

Mg(*) + 2HCl(aq)→MgCl2(aq) + H2(g)

(ii) sodium hydroxide is a neutralisation reaction. Sodium chloride salt and water are formed,Reaction between dilute hydrochloric acid and

NaOH(aq) + HCI(aq)→NaCl(aq) + H2O(l)

(iii) Dilute hydrochloric acid dissolves the CaCO3 and makes the shell soft.Egg shells are made of calcium carbonate, CaCO3.

CaCO3(*) + HCI(aq)→ CaCl2 + CO2(g) + H2O(l)

(Crushed egg shell)

...