# 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.

Nov 19, 2016

$\text{CaCO"_3}$ is the reactant in excess.
The amount of $\text{CaCO"_3}$ remaining is $\text{11.5 g}$.

#### Explanation:

Write a balanced equation.

$\text{CaCO"_3 + "2HCl}$$\rightarrow$$\text{CaCl"_2 + "CO"_2 + "H"_2"O}$

Use the balanced equation to determine the mole ratios between $\text{CaCO"_3}$ and $\text{CaCl"_2}$ and between $\text{HCl}$ and $\text{CaCl"_2}$ and between $\text{CaCO"_3}$ and $\text{HCl}$

Calcium carbonate and calcium chloride.

$\text{1 mol CaCO"_3/"1 mol CaCl"_2}$ and $\text{1 CaCl"_2/"1 mol CaCO"_3}$

Hydrochloric acid and calcium chloride.

$\text{2 mol HCl"/"1 CaCl"_2}$ and $\text{1 mol CaCl"_2/"2 HCl}$

Calcium carbonate and hydrochloric acid.

$\text{1 mol CaCO"_3/"2 mol HCl}$ and $\text{2 mol HCl"/"1 mol CaCO"_3}$

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

$28.0 \cancel{\text{g CaCO"_3xx(1 "mol CaCO"_3)/(100.1 cancel"g CaCO"_3)="0.280 mol CaCO"_3}}$

$12.0 \cancel{\text{g HCl"xx(1"mol HCl")/(36.5cancel"g HCl")="0.329 mol HCl}}$

Determine the mass of $\text{CaCl"_2}$ produced by each reactant by multiplying the moles of each reactant times the mole ratios with $\text{CaCl"_2}$ in the numerator. Then multiply the result by the molar mass of $\text{CaCl"_2}$ $\left(\text{111 g/mol}\right)$.

Calcium carbonate

$0.280 \cancel{\text{mol CaCO"_3xx(1cancel"mol CaCl"_2)/(1cancel"mol CaCO"_3)xx(111"g CaCl"_2)/(1cancel"mol CaCl"_2)="31.1 g CaCl"_2}}$

Hydrochloric acid

$0.329 \cancel{\text{mol HCl"xx(1cancel"mol CaCl"_2)/(2cancel"mol HCl")xx(111"g CaCl"_2)/(1cancel"mol CaCl"_2)="18.3 g CaCl"_2}}$

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 $\text{CaCO"_3}$ that reacted with the limiting reactant $\text{HCl}$.

$0.329 \cancel{\text{mol HCl"xx(1cancel"mol CaCO"_3)/(2cancel"mol HCl")xx(100.1"g CaCO"_3)/(1cancel"mol CaCO"_3)="16.5 g CaCO"_3color(white)(.)"reacted}}$

To determine the mass of $\text{CaCO"_3}$ that remains, subtract the mass that reacted from the starting mass.

$\text{28.0 g"-"16.5 g"="11.5 g}$