When 040 g of impure zinc reacted with an excess of hydrochl

When 0.40 g of impure zinc reacted with an excess of hydrochloric acid, 127 mL of hydrogen was

collected over water at 10.o

C. The external pressure was 737.7 Torr.

(a) What volume would the dry hydrogen occupy at 1.00 atm and 298 K?

(b) What amount (in moles) of H2 was collected?

(c) What is the percentage purity of the zinc, assuming that all the zinc present reacted

completely with HCl and that the impurities did not react with HCl to produce hydrogen?

The vapor pressure of water at 10.o

C is 9.21 Torr.

Solution

a) Using P1V1/T1 = P2V2/T2

(127 x 0.958)/283 = (1.00 x V2)/298

V2 = 128.11 mL

b) PV = nRT

0.958 x 0.127 = n x 0.082 x 283

n = 5.24 millimoles

c) moles of Zn = 0.40/65.38 = 6.12 millimoles

Zn + 2HCl = ZnCl2 + H2

actual moles of H2 which should be produced = 6.12 millimoles

% purity = (5.24/6.12) x 100 = 85.62%