If 37.1 mL AgNO3 solution reacts with excess potassium chloride solution to yield 1.56 g of AgCl precipitate, what is the molarity of silver ion in the original solution? AgCl = 143.3 g/mol Enter your answer in decimal format with three decimal places and no units.

Respuesta :

Answer:

Molarity of silver ion: 0.296 M

Explanation:

Reaction:

AgNO₃ + KCl → AgCl↓ + K⁺ + NO₃⁻

From the reaction, we know that the moles of AgCl produced will be the same as the moles of initial silver.

First, let´s calculate the number of moles of AgCl produced:

1.56 g AgCl was produced, that is, (1.56 g AgCl * 1 mol AgCl/143.3 g AgCl) 0.011 moles AgCl.

The moles of silver ion present in the original solution was 0.011 mol. Since this number of moles was present in a 37.1 ml solution, then, in 1000 ml:

moles of silver ion per liter = 1000 ml * 0.011 mol / 37.1 ml = 0.296 mol

Molarity of silver ion = 0.296 M