A sample of solid KClO_(3) (potassium chlorate) was heated in a test tube to obtain O_(2) according to the reaction 2KClO_(3) rarr 2KCl + 3O_(2) The oxygen gas was collected by downward displacement of water at 295K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mm of Hg at 300K. What is the partial pressure of the oxygen gas ?

A sample of solid KClO_(3) (potassium chlorate) was heated in a test t

1.	A sample of solid KClO_(3) (potassium chlorate) was heated in a test tube to obtain O_(2) according to the reaction 2KClO_(3) rarr 2KCl + 3O_(2) The oxygen gas was collected by downward displacement of water at 295K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mm of Hg at 300K. What is the partial pressure of the oxygen gas ?
Answer» Solution :`2KClO_(3(s)) RARR 2KCl_((s)) + 3O_(2 (g))` `P_("TOTAL")=772mm HG` `P_(H_(2)O) = 26.7 mm Hg` `P_("total") = P_(O_(2)) + P_(H_(2)O)` `:. P_(O_(2)) = P_("total") - P_(H_(2)O)` `P_(1) = 26.7 mm Hg T_(2) = 298K` `T_(1) = 300K, P_(2) = ?, (P_(1))/(T_(1)) = (P_(2))/(T_(2))` `P_(2) = ((P_(1))/(T_(1))) T_(2) = (26.7mm Hg)/(300 CANCEL(K)) xx 295 cancel(K)` `P_(2) = 26.26 mm Hg` `:. P_(O_(2)) = 772-26.26` =745.74 mm Hg