When 0.854 g of impure zinc reacted with an excess of hydrochloric acid, 244 mL

Question

When 0.854 g of impure zinc reacted with an excess of hydrochloric acid, 244 mL of hydrogen gas was collected over water at 10.0 °C. The external pressure was 714.96 Torr. The vapor pressure of water at various temperatures can be found in this table.

a. Calculate the volume the dry hydrogen would occupy at 1.00 atm and 298 K.

b. What amount, in moles, of H2 gas was collected at 10.0 °C?

c.Calculate the percentage purity of the zinc assuming that all of the zinc present reacted completely with HCl and that the impurities did not react with HCl to produce hydrogen gas.

Explanation / Answer

b) The vapor pressure of water at 100C is 9.21 torr.

Given,V=244 mL = 0.244 L , T = (10.0 + 273) K = 283 K

Now, PV = nRT

=> n = PV / RT

=> n = (714.96 torr - 9.21 torr) *0.244 L / (62.36367 Ltorr/Kmol) * 283 K

=> n =( 705.75 * 0.244 / 17648.91861) mol

=> n =( 172.203/17648.91861) mol

=> n =0.00976 mol of H2 gas was collected.

a) Given, T = 298 K and P= 1.00 atm , n = 0.00976 mol (calculated above)

Again, PV = nRT

=> V = nRT /P

=> V = (0.00976 mol) *(0.08206 L atm /mol K) *298 K / 1.00 atm

=> V = 0.238 L of dry hydrogen gas.

c) Zn + 2 HCl ---> ZnCl2  + H2

Now,

(0.00976 mol H2) x (1 mol Zn / 1 mol H2) x (65.38 g Zn /1 mol Zn ) / (0.854 g)

= (0.6381088/0.854 )

=0.7472 x 100%

= 74.72 % Zn

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