Question

93.What is the experimental yield (in grams) of the solid product when the percent yield is 84.33 % when 3.62 g of iron(III) chloride reacts in solution with excess sodium phosphate?

Answer 1

FeCl₃ (aq) + Na₃PO₄(aq)   ----> FePO₄(s) + 3NaCl(aq)

Molar mass of FeCl₃ = 162.5 g/mole

thus, moles of FeCl₃ in 3.62 g of it = mass/molar mass = 3.62/162.5 = 0.022

Hence moles of FePO₄ formed theoretically as per balanced reaction = moles of FeCl₃ reacting = 0.022

Molar mass of FePO₄ = 151 g/mole

Thus, theoretical mass of FePO₄ formed = moles*molar mass = 0.022*151 = 3.322 g

Now, % yield = (experimental mass/theoretical mass)*100

or, experimental mass = 2.801 g