Question

Question #8453e

Answer 1

`=530mL`

Explanation:

1. Write and balance the equation
   `2HgO(s)->2Hg(l)+O_2(g)`

2. Find the molar mass of `HgO` which relative atomic masses of the elements composing it are obtainable from the periodic table; i.e.,
   `HgO=216.6" g/mol"`

3. Find the moles of `HgO`, through molar conversion using the molar mass above as the conversion factor. Ensure that the units work out and the desired unit is attained.
   `=7.8cancel(gHgO)xx(1molHgO)/(216.6cancel(gHgO))`
   `=0.036molHgO`

4. Now, find the mole `O_2` by referring to the balanced equation where the molar ratio of `HgO` and `O_2` is obtainable from; i.e.,
   `=0.036cancel(molHgO)xx(1molO_2)/(2cancel(molHgO))`
   `=0.018molO_2`

5. Then, find the volume of `O_2`. Use the formula `PV=etaRT`, but prior to it make sure variables are up to its standard units as shown;
   `T="temperature"=150^oC+273=423K`
   `P="pressure"=120.3kPa`
   `R="gas constant"=(8.31446L*kPa)/(mol*K)`
   `eta="number of moles"=0.018mol`

6. Plug in values to the formula and cancel units as required to obtain the desired unit. Rearrange it and isolate the required variable; the volume (V);
   `PV=etaRT`
   `V=(etaRT)/(P)`
   `V=((0.018cancel(mol))((8.31446L*cancel(kPa))/cancel((mol*K)))(423cancel(K)))/(120.3cancel(kPa))`
   `V=0.5264L~~0.53L~~530ml`