calcium carbonate reacts with dilute hydrochloric acid, 1 g of calcium carbonate is added to 50 cm3 of 0.05 mol/dm^3 of hydrochloric acid which volume of carbon dioxide is made from this reaction
Answers
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
→
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
g CaCO
3
×
1
mol CaCO
3
100.1
g CaCO
3
=
0.280 mol CaCO
3
12.0
g HCl
×
1
mol HCl
36.5
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
mol CaCO
3
×
1
mol CaCl
2
1
mol CaCO
3
×
111
g CaCl
2
1
mol CaCl
2
=
31.1 g CaCl
2
Hydrochloric acid
0.329
mol HCl
×
1
mol CaCl
2
2
mol HCl
×
111
g CaCl
2
1
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
mol HCl
×
1
mol CaCO
3
2
mol HCl
×
100.1
g CaCO
3
1
mol CaCO
3
=
16.5 g CaCO
3
.
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