# What is the mass of mercury that can be prepared from 1.40 g of cobalt metal in the reaction Co(s) + HgCl_2(aq) -> CoCl_3(aq) + Hg(l)?

Jul 20, 2018

The mass of $\text{Hg}$ that can be produced is $\text{7.14 g}$.

#### Explanation:

Balanced equation

$\text{2Co(s) + 3HgCl"_2("aq")}$$\rightarrow$"2CoCl"_3("aq") + "3Hg("l")

Since the amount of $\text{HgCl"_2}$ is not given, it is assumed to be in excess.

To answer this question, there are three steps.

Determine mol $\text{Co}$ by multiplying it given mass by the inverse of its molar mass $\left(\text{58.993 g/mol}\right)$. This is the same as dividing by a fraction.

1.40color(red)cancel(color(black)("g Co"))xx(1"mol Co")/(58.993color(red)cancel(color(black)("g Co")))="0.0237 mol Co"

Determine mol $\text{Hg}$ by multiplying mol $\text{Co}$ by the mol ratio between them in the balanced equation, with mol $\text{Hg}$ in the numerator.

0.0237color(red)cancel(color(black)("mol Co"))xx(3"mol Hg")/(2color(red)cancel(color(black)("mol Co")))="0.0356 mol Hg"

Determine mass $\text{Hg}$ by multiplying mol $\text{Hg}$ by its molar mass $\left(\text{200.59 g/mol}\right)$.

0.0356color(red)cancel(color(black)("mol Hg"))xx(200.59"g Hg")/(1color(red)cancel(color(black)("mol Hg")))="7.14 g Hg"

The mass of $\text{Hg}$ that can be produced is $\text{7.14 g}$.

The three steps can be combined in one equation.

1.40color(red)cancel(color(black)("g Co"))xx(1color(red)cancel(color(black)("mol Co")))/(58.993color(red)cancel(color(black)("g Co")))xx(3color(red)cancel(color(black)("mol Hg")))/(2color(red)cancel(color(black)("mol Co")))xx(200.59"g Hg")/(1color(red)cancel(color(black)("mol Hg")))="7.14 g Hg"