Question

Equal length of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid ($HCl$) is added in test tube A, while acetic acid ($C{{H}_{3}}COOH$) is added in test tube B. In which test tube will the fizzing occur more vigorously?
(a)- A
(b)- B
(c)- Both
(d)- None

Answer 1

Magnesium is the element that belongs to group 2 of s-block and is placed on the left side of the periodic table. So, it is one of the very reactive elements and when it is treated with an acid it liberates hydrogen gas easily which causes fizziness.

Complete step by step answer:
Magnesium is the element which belongs to group 2 of s-block. Its atomic number is 12 and is placed on the left side of the periodic table. It is a metal and is one of the very reactive elements.
Since it is the element of group 2 it has a valency of +2. When it is treated with an acid it liberates hydrogen gas easily which causes fizziness.
So, in test tube A, magnesium ribbon and hydrochloric acid are added. As soon as the hydrochloric acid is added there is the liberation of hydrogen gas vigorously because hydrochloric acid is a strong acid. The reaction is given below:
$Mg+2HCl\to MgC{{l}_{2}}+{{H}_{2}}\uparrow$
In the test tube, B magnesium ribbon and acetic are added. When acetic acid is added there is liberation of hydrogen but not as vigorously as in hydrochloric acid because acetic acid is a weak acid. The reaction is given below:
$Mg+2C{{H}_{3}}COOH\to Mg{{(C{{H}_{3}}COO)}_{2}}+{{H}_{2}}\uparrow$

Therefore the correct answer is an option (a)- A.

Note: Magnesium reacts with oxygen to form oxides.
$2Mg+{{O}_{2}}\to 2MgO$
Only magnesium of its group reacts with nitrogen to form nitride.
$3Mg+{{N}_{2}}\to M{{g}_{3}}{{N}_{2}}$
When magnesium reacts with hydrogen it forms hydrides.
$Mg+{{H}_{2}}\to Mg{{H}_{2}}$