How many milliliters of oxygen gas at STP are released from heating 2.15 g of mercuric oxide? 2HgO(s)→2Hg(l)+O2(g)

Answer 1

Hello, Daijaih,

Thanks for adding the balanced equation. It tells us we'll get one mole of oxygen for every 2 moles of HgO. Not very productive, but the mercury is a lot of fun (before it was labelled hazardous). We can, and will, calculate the number of moles of O₂ based on the moles HgO we start with, Once we have the moles, we'll use the ideal gas law to calculate volume, at STP.

We start with only 2.5 g HgO, so I don't have high hopes we'll generate a large volume of O₂. Better to buy a tree.

Divide the grams HgO by it's molar mass to find moles HgO. I get 9.93x10^-3 moles HgO. Pitiful. The balanced equation tells us the molar ratio of O₂ to HgO is 1/2, or 0.5. That means we'll get only 1/2 the numebr of moles O₂, or 4.96x10^-3 moles O₂.

Now that we have moles O₂ we can use the ideal gas law to calculate volume at STP.

PV=nRT, R is the gas constant, noted in the table.

Rearrange to find V

V=nRT/P

P=1 atm (STP)

T = 273K (STP)

V = (4.96x10^-3 moles O₂)*(0.0821 L*atm/mole*K)(273K)/(1 atm)

V = 0.111 L, or 111ml.

I hope this helps,

Bob