Question

16.2gm of ZnO are reacted with 45.0ml of 6.0 M HCl solution. a) write a balanced equation for the reaction. b) which reactant is the limiting reagent? c) calculate the theoretical yield (gms) of salt produced. d) if 0.105 moles of salt are produced, what is the % yield?

Answer 1

Let; We have given; 16.2gm of ZnO are reacted with 45.0ml of 6.0M HCl solution;

a) Balanced equation for the reaction is;

ZnO_(s) + 2HCl_(aq) ZnCl_2(aq) + H₂O_(l)

b) Firstly calculating the number of moles of ZnO;

16.2 g of ZnO x(1 mole/81.38g) =0.1991 moles of ZnO

Also, Calculating number of moles of HCl =0.045Lx6.0M =0.27 moles of HCl

Now; Number of moles of ZnO required to react completely with HCl = 0.27 moles of HClx(1/2) =0.135 moles of ZnO

Similarly;  Number of moles of HCl required to react completely with ZnO = 0.1991 moles of ZnOx(2/1) =0.3981 moles of HCl

As moles of HCl required to react completely with ZnO are less; HCl is limiting Reagent.

c) Now; Calculating the theoretical yield(gms);

0.27 moles of HCl x( 1 mole of ZnCl₂ / 2 moles of HCl) =0.135 moles of ZnCl₂

Converting these moles to grams;

0.135 moles of ZnCl₂ x(136.286 g/1mole) =18.4g of ZnCl₂ [Theoretical yield]

d) We have given; Practical yield =0.105 molesx(136.286g/1mole) =14.3g

Now; Percentage yield = (14.3/18.4)x100

Percentage yield = 77.78%