Question

A sample of zinc metal is allowed to react completely with an excess of hydrochloric acid: Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(8) look GO The hydrogen gas produced is collected over water at 25.0°C. The volume of the gas is 8.90 L, and the atmospheric pressure is 0.951 atm. Calculate the amount of zinc metal in grams consumed in the reaction. rint rences (Vapor pressure of water at 25°C = 23.8 mmHg.) gZn

Answer 1

we can see from the given balanced equation that no.of moles of Zn reacted = no.of moles of H₂ gas produced

Given, T = 25^oC = 298 K, V =8.90 L ,

pressure due to water vapour = 23.8 mm of Hg = 23.8 $\div$ 760 atm = 0.031 atm

and total pressure = 0.951 atm = pressure due to water vapour + pressure due to H₂ gas

$\Rightarrow$ 0.951 atm = 0.031 + pressure due to H₂ gas

$\Rightarrow$ pressure due to H₂ gas = 0.920 atm.

Ideal gas equation: PV = nRT

for hydrogen gas, 0.920 $\times$ 8.90 = n $\times$ = 0.0821 $\times$ 298.0

n = 0.335 moles.

$\Rightarrow$ no.of moles of Zn reacted = no.of moles of H₂ gas produced = 0.335 moles.

atomic weight of Zn = 65.38 g/mol

weight of Zn metal = 0.335 $\times$ 65.38 = 21.9 g