Question

In Antoine Lavoisier’s classic experiment, mercuric oxide is heated in a sealed container. The solid red powder is changed into two products: silver liquid mercury and oxygen gas. If Lavoisier heated 100 grams of powdered mercuric oxide to produce 93 grams of liquid mercury, how much oxygen would be released?

A). 7 grams
B). 16 grams
C). 32 grams
D). 93 grams

Answer 1

A) 7 grams

Step-by-step explanation:

The chemical reaction is:

2HgO\rightarrow 2Hg + O2

Amount of HgO heated = 100 g

Molar mass of HgO = 216.59 g/mol

Molar mass of O2 = 32.00 g/mol

Moles of HgO heated =
`(Mass)/(Molar\ Mass) = (100)/(216.59) =0.462 moles`

Based on the reaction stoichiometry:

2 moles of HgO produces 1 mole of O2

Therefore, 0.462 moles of HgO would produce:

=
`(0.462\ moles\ HgO*1\ mole\ O2)/(2\ moles\ HgO) =0.231 moles`

Molar Mass of O2 = 32 g/mol

Mass of O2 =
`Moles*molar mass = 0.231\ moles * 32g/mol = 7.39 g`

Answer 2

After heating the compound, it is understood that what has been released is its water content. The remaining mercuric oxide, HgO, contains 7.39% oxygen. From the given above,
(93 grams) x (0.0739) = 6.87 grams
The obtained value is closest to 7. Thus, the answer is letter A.