Practice Exercise 9.5: Calculating Silver Nitrate Requirement

How many grams of silver nitrate are required to produce 0.25 mol of silver sulfide?

The balanced equation is: 2 AgNO₃ + H₂S → Ag₂S + 2 HNO₃

Answer:

To produce 0.25 mol of silver sulfide (Ag₂S) from silver nitrate (AgNO₃), you would need 84.935 grams of silver nitrate according to stoichiometry and molar mass calculations.

Stoichiometry is an important concept in chemistry that involves calculating reactant quantities based on a balanced chemical equation. In this case, we are looking to determine the amount of silver nitrate needed to produce a specific amount of silver sulfide.

From the balanced equation, we can see that 2 moles of silver nitrate (AgNO₃) are needed to produce 1 mole of silver sulfide (Ag₂S). Therefore, if we require 0.25 moles of Ag₂S, we will need twice that amount of AgNO₃, which is 0.5 moles.

The molar mass of silver nitrate (AgNO₃) is 169.87 g/mol. By multiplying the number of moles needed (0.5 moles) by the molar mass, we can obtain the mass of silver nitrate required, which is 84.935 grams.