Answer to Question #305509 in General Chemistry for Mcquay

Question

Pb(NO3)2 + 2 NaI = PbI2 + 2 NaNO3

6.Suppose in the reaction in which 100 grams of lead(II) nitrate were all used up and only 100 grams of lead(II)iodide were obtained in the experiment, what is the percent yield?

7.How many grams of the product were formed in no. 6 if the percent yield is 80%?

Answer 1

(i) Mass of lead nitrate = 100 g

Mass of I₂ produced= 100g

the give reaction isMass of lead nitrate = 1l00 g

$(PbNO_3)_2+2NaI \longrightarrow PbI_2 + 2NaNO_3$

Molar mass of lead nitrate =207.2 g/mol

Mole of lead nitrate = 0.483

Mole of lead iodide produced =0.483

Mass of lead iodide = 222.6 g

But real mass of lead iodide produces 100 g

So, mass percebtage = 44.9 %

(ii) for 80 % yield mass of leas iodide = 178.08 g