Write a balanced equation.
$"CaCO"_3 + "2HCl"$$rarr$$"CaCl"_2 + "CO"_2 + "H"_2"O"$
Use the balanced equation to determine the 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 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.
$"28.0 g"-"16.5 g"="11.5 g"$
