Question

In solution, 2 g of Calcium Chloride is reacted with 5 g of Lead (IV) Nitrate as illustrated in the

reaction below. A precipitate is formed and filtered out. The mass of the precipitate is 1.2 g.

Determine the percent yield.

Answer 1

$Pb(NO_3)_4(aq)+2CaCl_2(aq)→PbCl_4(s)+2Ca(NO_3)_2(aq)$

This is a precipitation reaction between two fully soluble salt solutions to form a new insoluble salt (lead (IV) chloride).

Word Equation:

lead (IV) nitrate + calcium chloride --> lead (IV) chloride + calcium nitrate

moles of CaCl₂ here=2/111

moles of lead nitrate here =5/455

here the limiting reagent is CaCl2hence the reaction will be governed by it

moles of precipitate formed here = 1/111

in mass = (1/111)*349=3.14g

but the yield found is 1.2 g

yield percent$=\frac{1.2}{3.14}\times100=38\%$