Question

What is the experimental yield (in g of precipitate) when 17.6 mL of a 0.7 M solution of iron(III) chloride is combined with 17.4 mL of a 0.613 M solution of silver nitrate at a 81.2% yield?

What is the theoretical yield (in g of precipitate) when 18.4 mL of a 0.73 M solution of sodium phosphate is combined with 18 mL of a 0.695 M solution of aluminum chloride?

Answer 1

1. Balanced reaction is:

FeCl₃(aq) + 3AgNO₃(aq) Fe(NO₃)₃(aq) + 3AgCl(s)

Molarity of FeCl₃ = 0.7 M = 0.7 mol/L

Volume of FeCl₃ = 17.6 mL = 0.0176 L

Moles of FeCl₃ = Molarity * Volume = 0.7 mol/L*0.0176 L = 0.01232 moles

Molarity of AgNO₃ = 0.613 M

VOlume of AgNO₃ = 17.4 mL = 0.0174 L

Moles of AgNO₃ = Molarity * Volume = 0.613M * 0.0174L = 0.01067 moles

From reaction,

1 mol of FeCl₃ reacts with 3 mol of AgNO₃

Thus, 0.01232 moles of FeCl₃ reacts with 3 mol* 0.01232 = 0.03696mol of AgNO₃

But we have

0.01067 moles of AgNO₃

THus, AgNO₃ is a limiting reagent.

Mole of AgCL formed = Moles of AgNO₃ reacted = 0.01067moles

Molar mass of AgCl = 143.32 g/mol

Mass of AgCl = Moles * Molar mass = 0.01067 mol* 143.32 g/mol = 1.529 g

Percentage yield = (Experimental yield/ Theortical yield) *100%

81.2% = (Experimental yield/1.529g)*100%

Experimental yield = 1.242 g

2. Reaction is:

Na₃PO₄(aq) + 3NH₄Cl(aq) 3NaCl(s) + (NH₄)₃PO₄(s)

Molarity of Na₃PO_{4 =} 0.73 M

Volume of Na₃PO₄ = 18.4 mL = 0.0184 L

Moles of Na₃PO₄ = 0.73 M * 0.0184 L = 0.01343 moles

Moles of NH₄Cl = 0.018L * 0.695 M = 0.01251 moles

From reaction,

3 moles of NH₄Cl reacts with 1 mol of Na₃PO₄

THus, 0.01251 moles of NH₄Cl reacts with 0.00417 mol of Na₃PO₄

THus, NH₄Cl is a limiting reagent.

So, Moles of (NH₄)₃PO₄ formed = 0.00417 moles

Molar mass of (NH₄)₃PO₄ = 149 g/mol

Mass of (NH₄)₃PO₄ = Moles * molar mass = 0.00417 moles * 149g/mol =0.6213g

Thus, theoretical yield is 0.6213 g