Question

Calculate the mass of Ag₂S produced when 125 mL of 0.200 M AgNO₃ is added to excess Na₂S solution. (Excess Na₂S means that you have more than you need - therefore, AgNO₃ is limiting.)

a. 0.150 g
b. 0.300 g
c. 0.450 g
d. 0.600 g

Answer 1

To calculate the mass of Ag₂S produced, we need to determine the limiting reactant between AgNO₃ and Na₂S. Since AgNO₃ is the limiting reactant, we can use stoichiometry to find the number of moles of Ag₂S produced. Finally, calculate the mass of Ag₂S by multiplying the moles of Ag₂S by the molar mass.

Step-by-step explanation:

To calculate the mass of Ag₂S produced, we need to determine the limiting reactant between AgNO₃ and Na₂S. Since AgNO₃ is the limiting reactant, we can use stoichiometry to find the number of moles of Ag₂S produced.

First, calculate the number of moles of AgNO₃ by multiplying the volume (125 mL) by the molarity (0.200 M) and converting to liters. Next, use the balanced equation to determine the stoichiometric ratio between AgNO₃ and Ag₂S. Since the ratio is 2:1, multiply the moles of AgNO₃ by 2 to find the moles of Ag₂S produced. Finally, calculate the mass of Ag₂S by multiplying the moles of Ag₂S by the molar mass.

Mass of Ag₂S produced = (2 * moles of AgNO₃) * molar mass of Ag₂S