An unknown amount of mercury (II) oxide was decomposed in the lab. Mercury metal
was formed and 6.60 L of oxygen gas was released at a pressure of 0.970 atm and
390.0 K. What was the initial weight of mercury oxide in the sample? (1 point)
1) 48.91 grams
2) 64.32 grams
3) 78.32 grams
4) 86.6 grams

The initial weight of mercury oxide in the sample is 4. 86.6 g (grams)

Explanation:

2HgO → 2Hg + O₂

From the given pressure, temperature and volume, we can find the number of moles of oxygen. Using the molar ratio, we can find the moles of HgO and then from that moles and molar mass we can find the mass of HgO.

P = 0.970 atm, T = 390 K , V= 6.60 L

PV = nRT

We need to find n as,

n = 0.2 moles of O₂

To find the moles of HgO,

The molar ratio is 1:2 for O₂ and HgO.

Moles of HgO =

Now we have to multiply the moles by the molar mass of HgO as,

0.4 moles ×216.59 g / mol = 86.6 g of HgO

Thus it is clear that the initial weight of mercury oxide is 86.6 g

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.