Part A How much potassium chlorate is needed to produce 21.0 mL of oxygen gas at

Question

Part A How much potassium chlorate is needed to produce 21.0 mL of oxygen gas at 680. mmHg Express the mass to three significant digits. and 24°C? Submit My Answers Give Up Part B of the wet gas were 715 mmHg If oxygen gas were collected over water at 22 °C and the total pressu Express the partial pressule in mm Hg to three significant digits what would be the partial pressure of the oxygen? Hints mmHg Submit My Answers Give Up Part C An oxide of nitrogen was found by elemental analysis to contain 30.4% nitrogen and 69.6% oxygen 123.0 g of ths gas were found to occupy 5.6 t at STP, what are the empirical and molecular formulas for this oxide of nitrogen? Express the empirical and molecular formulas for this oxide of nitrogen, separated by a comma

Explanation / Answer

A)

KClo3 required to form

KClO3 = KCl + 3/2O2

mol of O2 = ideal gas law

PV = nRT

n = PV/(RT) = 680*0.021/(62.4*(24+273)) = 0.0007705

now..

1.5 mol of O2 = 1 mol of KClO3

0.0007705mol of O2 = 0.0007705/1.5 = 0.00051366 mol of KClO3

mass = mol*MW = 0.00051366*122.55 = 0.06294 g of KClO3

B)

Ptotal = Pgas + Pvapor

Pgas = Ptotal - Pvapor

Pgas= 715 - 19.8 = 695.2 mm Hg

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