Question

What volume of oxygen gas is produced at step when a 20g sample of HgO(M=216.6g) is strongly heated

Answer 1

AAAaannndd the bot gets it wrong yet again!

since 1 mole occupies 22.4L at STP,
0.0461 mol O2 occupies .0461 * 22.4L = 1.03 L

Answer 2

The balanced chemical equation for the decomposition of HgO is:

2HgO(s) → 2Hg(l) + O2(g)

From this equation, we can see that for every 2 moles of HgO that decompose, 1 mole of O2 is produced.

First, we need to determine the number of moles of HgO in the 20g sample:

n(HgO) = m/M = 20g / 216.6g/mol = 0.0923 mol

Since 2 moles of HgO produce 1 mole of O2, we can calculate the number of moles of O2 produced:

n(O2) = 0.0923 mol HgO x (1 mol O2 / 2 mol HgO) = 0.0461 mol O2

Now, we can use the ideal gas law to calculate the volume of O2 produced:

PV = nRT

Assuming standard temperature and pressure (STP) of 273 K and 1 atm, respectively:

V = nRT/P = (0.0461 mol) x (0.0821 L·atm/mol·K) x (273 K) / (1 atm) = 1.18 L

Therefore, the volume of oxygen gas produced at the step when a 20g sample of HgO is strongly heated is approximately 1.18 L.