What volume of oxygen gas is produced when 70.7 g of mercury(II) oxide reacts completely according to the following reaction at 25 °C and 1 atm? mercury(II) oxide (s) mercury (l) + oxygen(g)

Answer 1

Hello, Julissa,

We need a balanced equation for this decomposition reaction.

2HgO = 2Hg + O₂

This tells us we get 1 mole of oxygen for every two moles of HgO. Jet's calculate the number of moles HgO in 70.7 grams by dividing by it's molar mass g/mole). The 0.326 moles of HgO will produce half that number of moles of O₂.

Now we need to convert that to volume under the specified conditions of temperature and pressue. Use the gas law:

PV = nRT

Make sure the temperature is entered as Kelvin.

Solve for V, volume:

V = nRT/P

V = (0.163 moles)*(0.0821 L*atm/mole*K)*(273+25K)

V = 3.99 L O₂

Bob