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)#?

Question

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)#?

Naomi Aronson · Accepted Answer

The mass of #"Hg"# that can be produced is #"7.14 g"#.

Equation in balance #"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. There are three steps to answering this question. 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"#. One equation can incorporate all three steps. #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"#

Jose Ayala · Answer

undefined To find the mass of mercury produced, you need to use stoichiometry. First, balance the equation:

Co(s) + 3HgCl₂(aq) -> CoCl₃(aq) + 3Hg(l)

Then, calculate the molar mass of cobalt and mercury.

Co: 1 atom x 58.93 g/mol = 58.93 g/mol 
Hg: 3 atoms x 200.59 g/mol = 601.77 g/mol

Next, calculate the moles of cobalt using its molar mass.

1.40 g Co x (1 mol Co / 58.93 g Co) = 0.0238 mol Co

According to the balanced equation, the molar ratio of cobalt to mercury is 1:3.

So, moles of Hg produced = 0.0238 mol Co x (3 mol Hg / 1 mol Co) = 0.0714 mol Hg

Finally, calculate the mass of mercury produced using its molar mass.

0.0714 mol Hg x (200.59 g Hg / 1 mol Hg) = 14.33 g Hg

Therefore, the mass of mercury produced is 14.33 grams.

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) -&gt; CoCl_3(aq) + Hg(l)#?

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)#?