Explain the action of the dilute hydrochloric acid on the following with chemical reaction:
i) Magnesium ribbon
ii) Sodium hydroxide
iii) Crushed egg shells
(i) The action of the dilute hydrochloric acid on the reaction with the magnesium ribbon is as follows:
When dilute hydrochloric acid reacts with magnesium ribbon, it forms magnesium chloride and hydrogen gas.
$$2HCl + Mg\rightarrow MgCl_2 + H_2$$
(ii) The action of the dilute hydrochloric acid on the reaction with the sodium hydroxide is as follows:
When dilute hydrochloric acid reacts with sodium hydroxide, it forms sodium chloride and water.
$$HCl + NaOH\rightarrow NaCl + H_2O$$
(iii) The action of the dilute hydrochloric acid on the reaction with the crushed egg cell$$(CaCO_3)$$ is as follows:
When dilute hydrochloric acid reacts with $$CaCO_3$$, it forms calcium chloride$$(CaCl_2)$$, carbon dioxide$$(CO_2)$$ and water$$(H_2O)$$
$$2HCl + CaCO_3\rightarrow CaCl_2 + CO_2 + H_2O$$