Limestone is mostly calcium carbonate, CaCO3, but also contains other minerals. When heated, the CaCO3 decomposes into CaO and CO2. A 1.605 g sample of limestone was heated and gave off 0.657 g of CO2.
(b) Determine the percentage mass of CaCO3 in the limestone.
Answers
Notice that the problem provides you with grams of carbon dioxide, but that the balanced chemical equation uses moles, so right from the start, you know that you must convert grams to moles or vice versa.
CaCO
3
(
s
)
+
2
HCl
(
a
q
)
→
CaCl
2
(
a
q
)
+
H
2
O
(
l
)
+
CO
2
(
g
)
↑
⏐
You know that
1
mole of calcium carbonate reacts to produce
1
mole of carbon dioxide. Use the molar masses of the two compounds to convert this mole ratio to a gram ratio.
You know that
M
M CaCO
3
=
100.1 g mol
−
1
M
M CO
2
=
44.01 g mol
−
1
This means that when
100.1 g
of calcium carbonate take part in the reaction, the equivalent of
1
mole of calcium carbonate, the reaction produces
44.01 g
of carbon dioxide, the equivalent of
1
mole of carbon dioxide.
You thus have
mole ratio
1 mole CaCO
3
1 mole CO
2
=
gram ratio
100.1 g CaCO
3
44.01 g CO
2
Use the gram ratio to determine how many grams of calcium carbonate were needed to produce
0.38 g
of carbon dioxide
0.38
g CO
2
⋅
100.1 g CaCO
3
44.01
g CO
2
=
0.864 g CaCO
3
To find the percentage of calcium carbonate in the sample of limestone, simply divide the mass of calcium carbonate by the total mass of the limestone and multiply the result by
100
%
% CaCO
3
=
0.864
g
1
g
×
100
%
=
86
%
−−−−−
I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the mass of limestone.
The percentage mass of CaCO₃ in the limestone is equal to 93.02%.
Given:
Mass of the Limestone sample = 1.605 g
Mass of the CO₂ formed = 0.657 g
To Find:
The percentage mass of CaCO₃ in the limestone sample.
Solution:
→ Limestone is a sedimentary rock primarily composed of calcium carbonate (CaCO₃), while the rest part being minerals such as the double carbonate of calcium and magnesium (dolomite).
→ When heated, the CaCO₃ decomposes into CaO and CO₂
The molar mass of CaCO₃ = 100 g
The molar mass of CO₂ = 44 g
The molar mass of CaO = 56 g
→ 1 mole of CaCO₃ decomposes to give 1 mole of CO₂ and 1 mole of CaO. In other words, we can say that 100 grams of CaCO₃ decomposes to give 44 grams of CO₂ and 56 grams of CaO.
CaCO₃ → CaO + CO₂
100 g 56 g 44 g
→ Hence 1.493 g of CaCO₃ is required to form 0.657 g of CO₂
→ This implies that a total of 1.493 g of CaCO₃ is present in the limestone sample.
→ This means that only 1.493 g of CaCO₃ is present in the 1.605 g of the limestone sample, with the rest part being other minerals.
Hence the percentage mass of CaCO₃ in the limestone comes out to be equal to 93.02%.
#SPJ2