in an experiment hydrochloric acid and reacted with calcium carbonate at STP and 80 cm cube of carbon dioxide was produced calculate the number of molecules of carbon dioxide given of
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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
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