Question

# A) Zinc metal reacts with excess hydrochloric acid to produce hydrogen gas according to the following...

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.

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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