Question

The following reaction forms 12.7 g of Ag(s): 2Ag2O(s)→4Ag(s)+O2(g) What total volume of gas forms if it is collected over water at a temperature of 25 ∘C and a total pressure of 760 mmHg ?

Answer 1

The given reaction is

  2Ag2O(s)→4Ag(s)+O2(g)

In this reaction 12.7 g of Ag(silver) has been formed, so now we need to determine the number of moles that will be formed when 12.7 g of Ag is produced.

So no of moles of Ag = 12.7/107.88 = 0.11123 moles

where 107.88 g is the molar mass of Ag.

Now from th reaction it is clear that for 4 moles of Ag it produces 1 mole of the O₂.

Thus the number of moles of  O₂ is 0.11123/4 = 0.02780 moles

Now as the gas is collected over water at temprature of 25 ∘C so we need to consder

Ptotal=Pwater+PO2 ⇒ PO2=Ptotal−Pwater

Pwater=vapour pressure of water

PO2= pressure of O2

So     PO2=760-26.69 =730.3 mm Hg

Now using the ideal ga sequation we have

PV=nRT

n=no of moles of O2=0.02780 moles

P=PO2=730.3 mm Hg

V = ?

T=273.15 + 25

V=nRT/P = 0.02780 x 0.0821 x 298.15 / (730.3/760)

V = 0.7081mL

thus total volume of gas forms if it is collected over water at a temperature of 25 ∘C and a total pressure of 760 mmHg is V = 0.7081mL