Question

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

Answer 1

Answer:

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

Explanation:

a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of $O_{2}$)

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

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

b) Lets assume that $O_{2}$ behaves ideally. Hence-

                                            PV=nRT

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

here P = 733 torr = $(733\times 0.001316)atm$ = 0.9646 atm

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

   So, $n=\frac{PV}{RT}$

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

                   = 0.0259 moles

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

Molar mass of $KClO_{3}$= 122.55 g

So mass of $KClO_{3}$ in sample = $(0.0173\times 122.55)g$

                                                                    = 2.12 g