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8. A sample of potassium chlorate (KCIO,) was heated in a test tube and decomposed 2KC?(s) 302 (g) + 2KCIO, (s) The oxygen was collected by the displacement of water at 22'C at a total pressure of 754 torr. The volume of the gas collected was 0.65L and the vapor pressure of water at 22°C is 21 torr. Calculate a) the partial pressure of O2 in the gas collected AND b) the mass of KCIO3 in the sample that was decomposed

Respuesta :

JannetPalos

Answer:

Partial pressure of [tex]O_{2}[/tex] in the gas was 733 torr and mass of [tex]KClO_{3}[/tex] in the sample was 2.12 g.

Explanation:

a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of [tex]O_{2}[/tex])

Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.

So, partial pressure of [tex]O_{2}[/tex]= (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr

b) Lets assume that [tex]O_{2}[/tex] behaves ideally. Hence-

                                            PV=nRT

where P is pressure of [tex]O_{2}[/tex], V is volume of [tex]O_{2}[/tex] , n is number of moles of [tex]O_{2}[/tex] , R is gas constant and T is temperature in kelvin

here P = 733 torr = [tex](733\times 0.001316)atm[/tex] = 0.9646 atm

        V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K

   So, [tex]n=\frac{PV}{RT}[/tex]

                   = [tex]\frac{(0.9646 atm)\times (0.65 L)}{(0.082 L.atm/(mol.K))\times (295 K)}[/tex]

                   = 0.0259 moles

As 3 moles of [tex]O_{2}[/tex] are produced from 2 moles of [tex]KClO_{3}[/tex] therefore 0.0259 moles of [tex]O_{2}[/tex] are produced from [tex](\frac{2\times 0.0259}{3})[/tex] moles or 0.0173 moles of [tex]KClO_{3}[/tex].

Molar mass of [tex]KClO_{3}[/tex]= 122.55 g

So mass of [tex]KClO_{3}[/tex] in sample = [tex](0.0173\times 122.55)g[/tex]

                                                                    = 2.12 g