I believe that it is petroleum ether.
15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.
<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>
Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2
<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO
</em>
Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>
After finding the oxidation states of atoms, you identify the half reactions (option c).
The half reactions are given by the change of the oxidation states of the atoms.
For example if Cu is in the left side with oxidation state 0 and in the other side with oxidation state 2+, then there you have a half reaction (oxidation reaction). And if you have O with oxidation state 0 in the left side and with oxidation state 2- in the right side, there you have other half reaction (reducing reaction).
Use formula: Initial Pressure x Initial Volume/Initial temperature = Final pressure x Final Volume/Final Temperature => 17.15L
1.3 x 10^21 atoms Na is 4.63230769231 mol Na
4.63230769231 mol Na * 23g/mol Na = 106.543076923g
