Given the following, how many g of calcium chloride will be produced when 28.0 g of calcium carbonate are combined with 12.0 g of hydrochloric acid? Which reactant is in excess and how many g of this reactant will remain after the reaction is complete?

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.

1 Answer
Nov 19, 2016

Answer:

#"CaCO"_3"# is the reactant in excess.
The amount of #"CaCO"_3"# remaining is #"11.5 g"#.

Explanation:

Write a balanced equation.

#"CaCO"_3 + "2HCl"##rarr##"CaCl"_2 + "CO"_2 + "H"_2"O"#

Use the balanced equation to determine the mole ratios between #"CaCO"_3"# and #"CaCl"_2"# and between #"HCl"# and #"CaCl"_2"# and between #"CaCO"_3"# and #"HCl"#

Calcium carbonate and calcium chloride.

#"1 mol CaCO"_3/"1 mol CaCl"_2"# and #"1 CaCl"_2/"1 mol CaCO"_3"#

Hydrochloric acid and calcium chloride.

#"2 mol HCl"/"1 CaCl"_2"# and #"1 mol CaCl"_2/"2 HCl"#

Calcium carbonate and hydrochloric acid.

#"1 mol CaCO"_3/"2 mol HCl"# and #"2 mol HCl"/"1 mol CaCO"_3"#

Determine the moles of each reactant by dividing the given masses by their molar masses.

#28.0 cancel"g CaCO"_3xx(1 "mol CaCO"_3)/(100.1 cancel"g CaCO"_3)="0.280 mol CaCO"_3"#

#12.0cancel"g HCl"xx(1"mol HCl")/(36.5cancel"g HCl")="0.329 mol HCl"#

Determine the mass of #"CaCl"_2"# produced by each reactant by multiplying the moles of each reactant times the mole ratios with #"CaCl"_2"# in the numerator. Then multiply the result by the molar mass of #"CaCl"_2"# #("111 g/mol")#.

Calcium carbonate

#0.280 cancel"mol CaCO"_3xx(1cancel"mol CaCl"_2)/(1cancel"mol CaCO"_3)xx(111"g CaCl"_2)/(1cancel"mol CaCl"_2)="31.1 g CaCl"_2"#

Hydrochloric acid

#0.329cancel"mol HCl"xx(1cancel"mol CaCl"_2)/(2cancel"mol HCl")xx(111"g CaCl"_2)/(1cancel"mol CaCl"_2)="18.3 g CaCl"_2"#

Since hydrochloric acid yields less calcium chloride than calcium carbonate, it is the limiting reactant .

Calcium carbonate is the excess reactant .

Determine the mass of #"CaCO"_3"# that reacted with the limiting reactant #"HCl"#.

#0.329 cancel"mol HCl"xx(1cancel"mol CaCO"_3)/(2cancel"mol HCl")xx(100.1"g CaCO"_3)/(1cancel"mol CaCO"_3)="16.5 g CaCO"_3color(white)(.)"reacted"#

To determine the mass of #"CaCO"_3"# that remains, subtract the mass that reacted from the starting mass.

#"28.0 g"-"16.5 g"="11.5 g"#