Question

. Consider this reaction: zinc metal reacts with hydrochloric acid to produce hydrogen and zinc chloride.

Zn + 2 HCl → ZnCl₂ + H₂

(a) If 0.30 mol of Zn is added to hydrochloric acid containing 0.52 mol of HCl, how many moles of H₂ are produced? (or what is the theoretical yield of H₂ produced in the reaction?)

(b) In the above procedure, if the actual yield of the reaction is 0.26 mol of H₂. What is the percent yield of the reaction?

(c ) If another piece of 12.46g sample of zinc reacts with excess HCl to form 0.298g of H₂. How pure is the zinc?

Answer 1

a)

Zn + 2 HCl   ---------->   ZnCl2 +    H2

1         2                             1             1

0.30     0.52

here limiting reagent = HCl .

2 mol HCl   ----------->   1 mol H2

0.52 mol HCl   ----------> ??

moles of H2 = 0.52 x 1 / 2 = 0.26

moles of H2 produced = 0.26

b)

actual yield = 0.26 mol

percent yield = (actual / theoretical) x 100

                      = (0.26 / 0.26) x 100

percent yield = 100 %

c)

mass of zn sample = 12.46 g

moles of Zn = 12.46 / 65.38 = 0.1905

Zn    + 2 HCl   ---------->   ZnCl2 +    H2

65.38 g     72.92 g                                   2.00 g

12.46 g                                                      ??

mass of H2 = 12.46 x 2 / 65.38

                    = 0.381 g

percent yield = (0.298 / 0.381) x 100

                      = 78.2 %