# Question 26c6d

Sep 13, 2015

The chemical reaction produced $\text{13.9 g}$ of liquid mercury.

#### Explanation:

From what I can understand from the information you given, it seems like you're dealing with a conservation of mass problem.

Mercury(II) oxide, $\text{HgO}$, can be decomposed by heating to liquid mercury, $\text{Hg}$, and oxygen gas, ${\text{O}}_{2}$.

The balanced chemical equation for this reaction looks like this

$2 {\text{HgO"_text((s]) -> 2"Hg"_text((l]) + "O}}_{\textrm{2 \left(g\right]}}$

Now, the idea here is that the mass of the reactant, which is mercury(II) oxide, must be equal to the mass of the products after the reaction has been completed.

SImply put, all the mercury(II) oxide will be converted into mercury liquid and oxygen gas.

All the atoms that started the reaction as part of the mercury(II) oxide will be found at the end of the reaction as either mercury liquid of oxygen gas.

This means that you have

${m}_{H g O} = {m}_{H g} + {m}_{{O}_{2}}$

The mass of liquid mercury that you can expect to find after the reaction is

${m}_{H g} = {m}_{H g O} - {m}_{{O}_{2}}$

m_(Hg) = "15.00 g" - "1.11 g" = color(green)("13.9 g")#

SIDE NOTE The answer is rounded to three sig figs, the number of sig figs you gave for the mass of oxygen gas.