Question

What quantity in moles of precipitate are formed when 25.0 mL of 1.00 M FeCl₃ reacts with 34.0 mL of 1.50 M AgNO₃ in the following chemical reaction? FeCl₃ (aq) + 3 AgNO₃ (aq) → 3 AgCl (s) + Fe(NO₃)₃ (aq)

Answer 1

0.0750 moles of precipitate (AgCl) are formed.

Step-by-step explanation:

To determine the quantity in moles of precipitate formed, we need to use the stoichiometry of the reaction. From the balanced chemical equation, we can see that the mole ratio between FeCl₃ and AgCl is 1:3. Therefore, 1 mole of FeCl₃ will produce 3 moles of AgCl.

First, let's calculate the moles of FeCl₃:

Moles of FeCl₃ = volume (in L) × molarity = (25.0 mL ÷ 1000 mL/L) × 1.00 M = 0.0250 moles

Using the mole ratio, we can now calculate the moles of AgCl:

Moles of AgCl = moles of FeCl₃ × (3 moles AgCl / 1 mole FeCl₃) = 0.0250 moles × 3 = 0.0750 moles

Therefore, 0.0750 moles of precipitate (AgCl) are formed.