Silver metal reacts with hydrogen sulfide (H₂S) to form solid silver sulfide (Ag₂S) according to the following balanced equation - Leaving Cert Chemistry - Question c - 2019

To find the number of moles (n), we use the formula:

$n = \frac{m}{M}$

where:

• m = mass of silver = 10.8 g
• M = molar mass of silver = 108 g/mol

Thus,

$n = \frac{10.8 \, \text{g}}{108 \, \text{g/mol}} = 0.1 \, \text{moles}$

From the balanced equation, the ratio of moles of Ag to moles of H₂ is 2:1, meaning 2 moles of silver react with 1 mole of hydrogen gas.

Based on the established ratio, using 0.1 moles of Ag:

$\text{Moles of H₂} = \frac{0.1 \, \text{moles Ag}}{2} = 0.05 \, \text{moles H₂}$

At standard temperature and pressure (s.t.p.), 1 mole of gas occupies 22.4 liters. Therefore:

$\text{Volume} = 0.05 \, \text{moles} \times 22.4 \, \text{liters/mole} = 1.12 \, \text{liters}$

To find the mass of Ag₂S formed, we first calculate the moles of Ag₂S produced:

From the balanced equation, 2 moles of Ag yield 1 mole of Ag₂S. Since we have 0.1 moles of Ag, we will form:

$\text{Moles of Ag₂S} = \frac{0.1}{2} = 0.05 \, \text{moles Ag₂S}$

Next, we find the molar mass of Ag₂S:

$\text{Molar mass of Ag₂S} = (2 \times 108) + (32) = 248 \, \text{g/mol}$

Thus, the mass of Ag₂S formed is:

$\text{Mass} = 0.05 \, \text{moles} \times 248 \, \text{g/mol} = 12.4 \, g$