Question

What volume of oxygen gas is produced when 53.4 g of mercury(II) oxide reacts completely according to the following reaction at 25 °C and 1 atm? mercury(II) oxide (s)>mercury (1) + oxygen(g) liters oxygen gas

Answer 1

2HgO_(s) ------------> Hg_(l) + O2_(g)

2 x 216.6 g HgO gives 1 mole O2

53.4 g HgO gives 53.4 x 1 / 2 x 216.6 = 53.4 / 433.2 = 0.123 moles O2

PV = nRT

V = nRT / P

V = 0.123 x 0.0821 x 298 / 1

V = 3.01 L O2