Chem 101 answers

Question

What quantity in moles of precipitate are formed when 25.0 mL of 1.00 M FeCl₃ reacts with 23.5 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

FeCl₃ + 3AgNO₃ ==> Fe(NO₃)₃ + 3AgCl(s) ... balanced equation

Find the limiting reactant. An easy way is to divided the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less is limiting.

For FeCl₃: 25.0 ml x 1 L / 1000 ml x 1.00 mol/L = 0.025 mols FeCl₃ (÷1->0.025)

For AgNO₃: 23.5 ml x 1 L / 1000 ml x 1.50 mol/L = 0.0353 mols AgNO₃ (÷3->0.012) = LIMITING

Use the moles of the limiting reactant to find moles of the product, AgCl.

0.0353 mol AgNO₃ x 3 mol AgCl / mol AgNO₃ = 0.106 moles AgCl (3 sig. figs.)