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?

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.

1 Answer

Impact of this question

121663 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-19T20:57:39

${CaCO}_{3}$ $"CaCO"_3"$ is the reactant in excess.
The amount of ${CaCO}_{3}$ $"CaCO"_3"$ remaining is $11.5 g$ $"11.5 g"$ .

Explanation:

Write a balanced equation.

${CaCO}_{3} + 2HCl$ $"CaCO"_3 + "2HCl"$ $\to$ $rarr$ ${CaCl}_{2} + {CO}_{2} + H_{2} O$ $"CaCl"_2 + "CO"_2 + "H"_2"O"$

Use the balanced equation to determine the mole ratios between ${CaCO}_{3}$ $"CaCO"_3"$ and ${CaCl}_{2}$ $"CaCl"_2"$ and between $HCl$ $"HCl"$ and ${CaCl}_{2}$ $"CaCl"_2"$ and between ${CaCO}_{3}$ $"CaCO"_3"$ and $HCl$ $"HCl"$

Calcium carbonate and calcium chloride.

$\frac{{1 mol CaCO}_{3}}{{1 mol CaCl}_{2}}$ $"1 mol CaCO"_3/"1 mol CaCl"_2"$ and $\frac{{1 CaCl}_{2}}{{1 mol CaCO}_{3}}$ $"1 CaCl"_2/"1 mol CaCO"_3"$

Hydrochloric acid and calcium chloride.

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

Calcium carbonate and hydrochloric acid.

$\frac{{1 mol CaCO}_{3}}{2 mol HCl}$ $"1 mol CaCO"_3/"2 mol HCl"$ and $\frac{2 mol HCl}{{1 mol CaCO}_{3}}$ $"2 mol HCl"/"1 mol CaCO"_3"$

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.

$"28.0 g"-"16.5 g"="11.5 g"$

Science

Math

Humanities

... and beyond

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

Explanation:

Related questions

Impact of this question