Question

What are zinc granules? Explain with examples.

Answer 1

In order to answer this question, first we will explain the definition or the brief explanation of zinc granules and then we will also write the chemical reaction of zinc granules with dilute sulphuric acid. And then we will explain the reaction.

Complete answer:
Zinc granules are unevenly shaped particles of zinc that are formed by pouring molten zinc into water. Zinc granules have a greater particle size as compared to zinc powder. Zinc is employed in the manufacture of hydrogen gas in the form of granules due to its enormous surface area. Both zinc granules and zinc have the same physical properties, although chemical attributes differ only slightly.
In the laboratory preparation of hydrogen, zinc granules are favoured over pure zinc because the impurity contained in granulated zinc is copper, which has a catalysing effect and speeds up the reaction rate.
Example-
The reaction between zinc metal and $HCl$ :
$Zn + HCl \to {H_2} + ZnC{l_2}$
Zinc metal displaces hydrogen to generate hydrogen gas and zinc chloride, a salt, in a single replacement reaction.
The solid crystals of zinc are known as zinc granules. Zinc reacts easily with the acid to produce hydrogen bubbles.
$Zn + HCl \to {H_2} + ZnC{l_2}$ is a reaction involving zinc and hydrochloric acid.
It's a single replacement reaction in which hydrogen is replaced by zinc metal, resulting in hydrogen gas and zinc chloride as a salt. Zinc interacts quickly with the acid to produce hydrogen bubbles.
When zinc is mixed with dilute sulphuric acid, hydrogen gas is released, resulting in zinc chloride. The reaction is exothermic, which means it produces heat.

Note:
The complex ion ${\left[ {Zn{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$ is present in aqueous solution as the $Zn(II)$ ion. In moist air, zinc reacts with oxygen. The metal burns in the air to generate $Zn(II)$ oxide, a substance that turns yellow when heated for a long time. Zinc reacts with gaseous bromine $(B{r_2})$ and iodine $({I_2})$ to form $({I_2})$ .