Question

Consider this reaction, which occurs in the atmosphere and contributes to photochemical smog:

ZnO(s) + H₂O(l) →Zn(OH)₂(aq)

If there is 16.9 g ZnO and excess H₂O present, the reaction yields 18.5 g Zn(OH)₂. Calculate the percent yield for the reaction.

Metallic zinc reacts with aqueous HCl. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) What volume of 2.60 M HCl, in milliliters, is required to convert 12.6 g of Zn completely to products? mL HCl

Answer 1

part1) ZnO(s) + H2O(l) →Zn(OH)2(aq)

according to balanced reaction

81.38 g ZnO gives 99.424 g Zn(OH)2

16.9 g ZnO gives 16.9 x 99.424 / 81.38 = 20.65 g Zn(OH)2

therotical yield = 20.65 g

actual yield = 18.5 g

% yield = (actual yield / therotical yield) x 100

% yield = (18.5 / 20.65) x 100

% yield = 89.60 %

part 2) Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

according to balanced reaction

65.38 g Zn reacts with 2 moles HCl

12.6 g Zn reacts with 12.6 x 2 / 65.38 = 0.38 moles HCl

moles = M x V in L

V = moles /M

V = 0.38 / 2.60

V = 0.146 L

V = 146 mL HCl required