How many grams of oxygen can be prepared by the decomposition of 12 grams of mercury oxide
1 answer:
Taking into account the reaction stoichiometry, 0.886 grams of O₂ can be prepared by the decomposition of 12 grams of mercury oxide.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction of the decomposition of mercury oxide is:
2 HgO → 2 Hg + O₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- HgO: 2 moles
- Hg: 2 moles
- O₂: 1 mole
The molar mass of the compounds is:
- HgO: 216.59 g/mole
- Hg: 200.59 g/mole
- O₂: 32 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
HgO: 2 moles ×216.59 g/mole= 433.18 grams
Hg: 2 moles ×200.59 g/mole= 401.18 grams
O₂: 1 mole ×32 g/mole= 32 grams
<h3>Mass of oxygen formed</h3>The following rule of three can be applied: if by reaction stoichiometry 433.18 grams of HgO form 32 grams of O₂, 12 grams of HgO form how much mass of O₂?
<u><em>mass of O₂= 0.886 grams</em></u>
Then, 0.886 grams of O₂ can be prepared by the decomposition of 12 grams of mercury oxide.
Learn more about the reaction stoichiometry:
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