Question

The minimum quantity in grams of ${H}_{2}S$ needed to precipitate 63.5 g of ${Cu}^{2+}$ will be nearly:
a.) 63.5 g
b.) 31.75 g
c.) 34 g
d.) 20 g

Answer 1

Hint: The concept involved in the above reaction is of mole, also known as mole concept. It can be used to find the masses or the moles of the substance used or formed in the reaction.

Complete step by step answer:
- Mole concept is a very convenient method which is used for expressing the amount of substance. The measurement can be down in two parts - numerical magnitude and the units that the magnitude is expressed in.
- Now, let us apply this concept in the given question.
In the reaction, ${H}_{2}S$ reacts with ${Cu}^{2+}$, which results in the precipitation of $CuS$. The reaction involved is given below.
$ { H }_{ 2 }S\quad +\quad { Cu }^{ +2 }\quad \longrightarrow \quad \underset { Precipitate }{ CuS\downarrow }$
Now, to precipitate 1 ion of ${Cu}^{2+}$, we require 1 molecule of ${H}_{2}S$. If we talk in terms of mole, 1 mole of ${H}_{2}S$ is required to precipitate 1 mole of ${Cu}^{2+}$ ion.
Mass of 1 mole of ${Cu}^{2+}$ = 63.5 g
Mass of 1 mole of ${H}_{2}S$ = 2 + 32 = 34 g
Therefore, 63.5 g of ${Cu}^{2+}$ reacts with 34 g of ${H}_{2}S$.
Hence, the mass of ${H}_{2}S$ needed in grams is 34 g.
So, the correct answer is “Option C”.

Note: There is a difference between the molecular mass and the molar mass. Molecular mass is the sum of the atomic masses of the individual atoms in a molecule, whereas, molar mass is the total mass of one mole of the substance.