Question

We wish to determine the moles of solid AgCl formed when 50.0 ml of 0.250 m AgNO₃ reacts wiht excess MgCl₂ according to the equation below. 2AgNO₃(aq) + MgCl₂(aq) → 2AgCl(s) + Mg(NO₃)₂(aq) in the previous step, you determined 0.0125 mol agno3 react. how many mole of AgCl form during the reaction?

Answer 1

0.0125 mol of AgCl form when 0.0125 mol of AgNO₃ react with excess MgCl₂, as the reaction has a 1:1 molar ratio between AgNO₃ and AgCl.

Step-by-step explanation:

The amount of moles of AgCl formed when 50.0 mL of 0.250 M AgNO₃ reacts with excess MgCl₂ can be determined using stoichiometry. According to the reaction 2AgNO₃(aq) + MgCl₂(aq) → 2AgCl(s) + Mg(NO₃)₂(aq), the ratio of AgNO₃ to AgCl is 1:1. Therefore, if 0.0125 mol of AgNO₃ reacts, an equal amount of moles of AgCl will form.

Using the stoichiometry of the reaction:
2AgNO₃(aq) + MgCl₂(aq) → 2AgCl(s) + Mg(NO₃)₂(aq),
for every mole of AgNO₃ that reacts, one mole of AgCl is produced. Since 0.0125 mol of AgNO₃ are reacted, 0.0125 mol of AgCl will form as well.