**Balanced equation**

#"2Co(s) + 3HgCl"_2("aq")"##rarr##"2CoCl"_3("aq") + "3Hg("l")#

Since the amount of #"HgCl"_2"# is not given, it is assumed to be in excess.

To answer this question, there are three steps.

Determine mol #"Co"# by multiplying it given mass by the inverse of its molar mass #("58.993 g/mol")#. 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 #"Hg"# by multiplying mol #"Co"# by the mol ratio between them in the balanced equation, with mol #"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 #"Hg"# by multiplying mol #"Hg"# by its molar mass #("200.59 g/mol")#.

#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 #"Hg"# that can be produced is #"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"#