Question

zinc metal reacts with excess hydrochloric acid to produce hydrogen gas according to the following equation: Zn(s) + 2HCl(aq)ZnCl2(aq) + H2(g) The product gas, H2, is collected over water at a temperature of 20 °C and a pressure of 745 mm Hg. If the wet H2 gas formed occupies a volume of 9.32 L, the number of moles of Zn reacted was __mol. The vapor pressure of water is 17.5 mm Hg at 20 °C.

Answer 1

The vapor pressure of water is 17.5 mm Hg at 20 °C.

The vapor pressure of wet hydrogen gas is 745 mm Hg

The vapor pressure of dry hydrogen gas mm Hg

Number of moles of hydrogen gas

$n = \dfrac {PV}{RT} = \dfrac { \dfrac { \text { 727.5 mm Hg} }{ \text { 760 mm Hg / atm }} \times \text { 9.32 L } }{ \text { 0.08206 L atm / mol K} \times \text { 293.15 K }}$

The number of moles of Zn reacted = the number of moles of hydrogen formed = 0.371