How many grams of oxygen can be prepared by the decomposition of 12
grams of mercury (ll) oxide?
The balanced chemical equation for the decomposition of mercury (II) oxide is:
2 HgO → 2 Hg + O2
According to the equation, for every 2 moles of HgO, we produce 1 mole of O2.
To find the number of moles of HgO in 12 grams, we need to use the molar mass of HgO. The molar mass of HgO is 200.59 g/mol (1 Hg atom = 200.59 g/mol, 1 O atom = 16 g/mol).
Number of moles of HgO = mass of HgO / molar mass of HgO
Number of moles of HgO = 12 g / 200.59 g/mol
Number of moles of HgO ≈ 0.0599 mol
According to the balanced equation, for every 2 moles of HgO, we produce 1 mole of O2. Therefore, the number of moles of O2 produced is half the number of moles of HgO.
Number of moles of O2 = 0.0599 mol / 2
Number of moles of O2 ≈ 0.02995 mol
Finally, to calculate the mass of O2 produced, we multiply the number of moles of O2 by its molar mass.
Mass of O2 = number of moles of O2 * molar mass of O2
Mass of O2 = 0.02995 mol * 32 g/mol
Mass of O2 ≈ 0.9584 g
Therefore, approximately 0.9584 grams of oxygen can be prepared by the decomposition of 12 grams of mercury (II) oxide.