What mass of precipitate can be produced when 120.0 mL of .130 M barium chloride and 160.0 mL of 0.140 M iron(Ill) sulfate are mixed?

2 Answers
Jul 31, 2017

Answer:

Approx. #3.6*g# #"barium sulfate"# will precipitate.

Explanation:

#Fe_2(SO_4)_3(aq) + 3BaCl_2(aq)rarr2FeCl_3(aq) + 3BaSO_4(s)darr#

#"Moles of barium ion"=120.0xx10^-3*Lxx0.130*mol*L^-1=0.0156*mol#.

#"Moles of sulfate ion"=3xx160.0xx10^-3*Lxx0.140*mol*L^-1=0.0672*mol#.

Sulfate ion is in excess, and thus #BaSO_4(s)# should precipitate quantitatively.....

We get #0.0156*molxx233.43*g*mol^-1=??*g#

Jul 31, 2017

Answer:

#3.64# #"g BaSO"_4#

Explanation:

We're asked to find the theoretical yield of the precipitate of a reaction, given the reactants and their amounts.

We'll first write a balanced chemical equation for this reaction:

#3"BaCl"_2(aq) + "Fe"_2"(SO"_4")"_3(aq) rarr 3"BaSO"_4(s) + 2"FeCl"_3(aq)#

What we need to do is find the limiting reactant, by first finding the moles of each reactant present (via the molarity equation) and then dividing the mole value by their respective coefficients; whichever reactant has the lower value is limiting.

#"mol BaCl"_2 = ("molarity")("L soln") = (0.130M)overbrace((0.1200color(white)(l)"L"))^"convert volume to L"#

#= ul(0.0156color(white)(l)"mol BaCl"_2#

#"mol Fe"_2"(SO"_4")"_3 = (0.140M)overbrace((0.1600color(white)(l)"L"))^"convert volume to L"#

#= ul(0.0224color(white)(l)"mol Fe"_2"(SO"_4")"_3#

Divide by respective coefficients to find limiting reactant:

#(0.0156color(white)(l)"mol BaCl"_2)/(3color(white)(l)"(coefficient)") = ul(0.00520#

#(0.0224color(white)(l)"mol Fe"_2"(SO"_4")"_3)/(1color(white)(l)"(coefficient)") = 0.0224#

Barium chloride is thus the limiting reactant.

Now we use the mole value of #"BaCl"_2# (#0.0156# #"mol"#) and the coefficients of the chemical equation to find the relative number of moles of #"BaSO"_4# (the precipitate) that can form:

#0.0156cancel("mol BaCl"_2)((3color(white)(l)"mol BaSO"_4)/(3cancel("mol BaCl"_2))) = color(red)(0.0156color(white)(l)"mol BaSO"_4#

Finally, we use the molar mass of #"BaSO"_4# (#233.39# #"g/mol"#) to find the mass in grams:

#color(red)(0.0156)cancel(color(red)("mol BaSO"_4))((233.39color(white)(l)"g BaSO"_4)/(1cancel("mol BaSO"_4))) = color(blue)(ul(3.64color(white)(l)"g BaSO"_4#