Question

How many grams of oxygen can be prepared by the decomposition of 12 grams of mercury oxide​

Answer 1

Taking into account the reaction stoichiometry, 0.886 grams of O₂ can be prepared by the decomposition of 12 grams of mercury oxide.

Reaction stoichiometry

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

Mass of oxygen formed

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?

$mass of O_{2} =\frac{12 grams of HgOx 32 grams of O_{2}}{433.18 grams of HgO}$

mass of O₂= 0.886 grams

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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