Question

Chlorine is prepared in the laboratory by treating manganese dioxide $\left( {Mn{O_2}} \right)$ with aqueous hydrochloric acid according to reaction.
$MnO2{\text{ }} + {\text{ }}4HCl \to {\text{ }}MnCl2 + Cl2 + 2H2O$
How many grams of $HCl$will react with 5 gm. of$Mn{O_2}$?

Answer 1

Hint-We have to determine the number of moles of$HCl$ required completing the reaction. We can use the values of the number of moles to determine the mass of the compound.

Complete step by step answer:
From the given chemical equation of the reaction is
$MnO2{\text{ }} + {\text{ }}4HCl \to {\text{ }}MnCl2 + Cl2 + 2H2O$
So, in this reaction we can understand that the 1 mole of $Mn{O_2}$ reacts with 4 moles of $HCl$
Now,
Mass of $Mn{O_2}$ = atomic mass of Manganese + 2(atomic mass of oxygen)
Therefore, we can write the mass of $Mn{O_2}$ is =$55{\text{ }} + 2\left( {16} \right)$
$= {\text{ }}87{\text{ }}gms/mole$
This means, 1 mole of $Mn{O_2}$ comprises 87 gm of MnO2.
Similarly,
Mass of $HCl$ = atomic mass of Hydrogen + atomic mass of Chlorine
Therefore, we can write as mass of $HCl$ = $1{\text{ }} + {\text{ }}35.45$
$= 36.46{\text{ }}gms/mole$
This means, 1 mole of $HCl$ comprises a mass of 36.46 gm of MnO2.
So, 4 moles of $HCl$ = 4 ×36.46 gm. = 145.84 gm of $HCl$
Thus, according to given chemical equation,
145.84 gm. of $HCl$ reacts with 87 gm. of $Mn{O_2}$ to accomplish the reaction.
So now we can find the mass of $HCl$ reacts with 5gm of $Mn{O_2}$ to accomplish the reaction.
$\therefore$ Mass of $HCl$ (gm.) required = $\dfrac{{145.84 \times 5}}{{87}} =$ 8.38 gm.
Hence, the answer is, 8.38 gm. of $HCl$ will require to react with 5 gm. of $Mn{O_2}$

Note: We should have knowledge of atomic masses of the elements in order to calculate the mass of compounds in the given chemical reaction and also about the calculating number of moles based on the chemical equation given.