Question

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.

Answer 1

`"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"`