CaCO3+2HCl---> CaCl2+H2O+CO2
The volume of CO2 gas formed when 2.5g calcium carbonate are dissolved in excess hydrochloric acid at 0 °C and 1 atm pressure is
[1 mole of any gas at 0 °C and 1 atm pressure
occupies 22.414 1 volume).
(a) 1.121
(b) 56.0L
(C) 0.28 L
(d) 0.561
Pl explain also
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Answers
Answer: correct answer you have not given
the correct answer is 0.56 L
Explanation:
CaCO
3
+2HCl→CaCl
2
+H
2
O+CO
2
100 g 44 g
1 mole 1 mole
22.4 L 22.4 L
We know that 1 mole of any gas at 0
o
C and 1 atm pressure occupies 22.4 L volume.
So, 100 g CaCO
3
forms 22.4 L of CO
2
Hence, 2.5 g CaCO
3
will form =
100
2.5×22.4
=
100
56
L of CO
2
=0.56 L of CO
2