Question

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

Answer 1

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

Step-by-step 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=(P V)/(R T)=(0.970 * 6.60)/(0.08206 * 390)=0.2 \text { moles }`

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 =
`\frac{0.2 \text { moles of } o_(2) * 2 \text { moles of } \mathrm{HgO}}{1 \text { mol of } \mathrm{O}_(2)}=0.4 \text { moles of } \mathrm{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