
When magnesium reacts with dilute \[HCl\] , the gas formed is:
A. chlorine
B. hydrogen
C. hydrogen chloride gas
D. none of these
Answer
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Hint: This is an example of single replacement reaction. Here oxidation and reduction takes place between magnesium and \[HCl\]
Complete step by step answer:
Magnesium in an element in the periodic table with atomic number \[12\]. It belongs to alkaline earth metals. Magnesium is highly electropositive in nature. It has the tendency to donate electrons from the valence shell of the atom.
The electronic configuration of magnesium is:
$Mg:1{s^2}2{s^2}2{p^6}3{s^2}$
The valence shell of magnesium is third. So when it is treated with an electronegative species, it readily loses electrons to produce an ionic compound. The electron present is s orbitals of magnesium overlap with the p orbitals of the electronegative atom.
In this case the magnesium metal is treated with \[HCl\] which will result in the formation of magnesium chloride and hydrogen gas. The corresponding balanced equation is:
$Mg(s) + 2HCl(l) \to MgC{l_2}(s) + {H_2}(g)$
The oxidation state of magnesium in \[MgC{l_2}\] is \[ + 2\] obtained by losing two \[3s\] electrons. The balanced equation for the reaction indicated a molecule of hydrogen is also released during the reaction.
So when magnesium is treated with dilute \[HCl\] , the gas formed is hydrogen i.e. option B is the correct answer.
Chlorine gas is prepared by treating oxide of magnesium like magnesium dioxide with conc. hydrochloric acid. The corresponding chemical equation is:
$Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O$
Hydrogen chloride gas is prepared in the laboratory by treating sodium chloride and sulfuric acid.
$NaCl + {H_2}S{O_4} \to NaHS{O_4} + HCl$
$NaHS{O_4} + NaCl \to N{a_2}S{O_4} + HCl$
option B is the correct answer.
Note:
Hydrogen gas is prepared in the laboratory by treating zinc with dil. \[HCl\] or dil.\[{H_2}S{O_4}\] . Sodium like magnesium is used to prepare dry solvents in the laboratory. The water present in organic solvent reacts with sodium releasing hydrogen gas thereby making the solvent water free.
Complete step by step answer:
Magnesium in an element in the periodic table with atomic number \[12\]. It belongs to alkaline earth metals. Magnesium is highly electropositive in nature. It has the tendency to donate electrons from the valence shell of the atom.
The electronic configuration of magnesium is:
$Mg:1{s^2}2{s^2}2{p^6}3{s^2}$
The valence shell of magnesium is third. So when it is treated with an electronegative species, it readily loses electrons to produce an ionic compound. The electron present is s orbitals of magnesium overlap with the p orbitals of the electronegative atom.
In this case the magnesium metal is treated with \[HCl\] which will result in the formation of magnesium chloride and hydrogen gas. The corresponding balanced equation is:
$Mg(s) + 2HCl(l) \to MgC{l_2}(s) + {H_2}(g)$
The oxidation state of magnesium in \[MgC{l_2}\] is \[ + 2\] obtained by losing two \[3s\] electrons. The balanced equation for the reaction indicated a molecule of hydrogen is also released during the reaction.
So when magnesium is treated with dilute \[HCl\] , the gas formed is hydrogen i.e. option B is the correct answer.
Chlorine gas is prepared by treating oxide of magnesium like magnesium dioxide with conc. hydrochloric acid. The corresponding chemical equation is:
$Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O$
Hydrogen chloride gas is prepared in the laboratory by treating sodium chloride and sulfuric acid.
$NaCl + {H_2}S{O_4} \to NaHS{O_4} + HCl$
$NaHS{O_4} + NaCl \to N{a_2}S{O_4} + HCl$
option B is the correct answer.
Note:
Hydrogen gas is prepared in the laboratory by treating zinc with dil. \[HCl\] or dil.\[{H_2}S{O_4}\] . Sodium like magnesium is used to prepare dry solvents in the laboratory. The water present in organic solvent reacts with sodium releasing hydrogen gas thereby making the solvent water free.
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