60 g of limestone on heating produced 22 gm of co2.The percentage of caco3 in limestone is
Answers
The answer of this question is 83.3% the solution is in below photos
Answer:
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.