Question

A)

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 757 mm Hg. If the wet H₂ gas formed occupies a volume of 7.17L, the number of moles of Zn reacted was ___________ mol. The vapor pressure of water is 17.5 mm Hg at 20 °C.

B) Oxygen gas can be prepared by heating potassium chlorate according to the following equation:

2KClO₃(s)-----------> 2KCl(s) + 3O₂(g)

The product gas, O₂, is collected over water at a temperature of 25 °C and a pressure of 748 mm Hg. If the wet O₂ gas formed occupies a volume of 9.15L, the number of moles of KClO₃ reacted was __________ mol. The vapor pressure of water is 23.8 mm Hg at 25 °C.

Answer 1

We have to use PV=nRT in both cases

Q1

Here according to the reaction moles of Zn reacted=moles of H2 produced and we can calculate moles of H2 using the above relation we know P=757-17.5=739.5 mm of Hg T=20 C =293.15 K V=7.17 L

hence we can calculate number of moles of H2  

n=0.29 moles

Hence mass of Zn reacted =0.29*65.38=18.95 g

Q2

Here according to the reaction moles of KClO3 reacted=(moles of O2 produced)/1.5 and we can calculate moles of O2 using the above relation we know P=748-23.8=724.2 mm of Hg T=25 C =298.15 K V=9.15 L

hence we can calculate number of moles of O2

n=0.35 moles

Hence mass of KClO3 reacted =(0.35*122.55)/1.5=29.10 g

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