Given:
Mass of
CaCO3 = 10 g
Volume of HCl = 1 L
Molarity of HCl = 0.1 M
We have to find the volume of CO2 produced.
Chemical reaction of calcium carbonate:
CaCO3 with HCl is given as：
CaCO3 + 2HCl →
CaCl2+CO2+H2O ... (i)
Molecular mas of
CaCO3 = 40 + 12 + 3 × 16
= 100 g
Now, number of moles is given as :
Number of moles =
... (i)
Thus, no. of moles of
CaCO3 =
= 0.1 moles
Now, molarity is given as
Molarity =
| No.ofmolessolute |
| Volumeofsolution(inmL) |
... (ii)
Substituting the values of molarity of HCl and volume of HCl in equation (i), we get
0.1 =
Thus, number of moles of HCl = 0.1
From reaction (i), 1 mole of
CaCO3 reacts with 2 moles of HCl to produce 1 mole of
CO2 Moles of HCl required is double the moles of
CaCO3.
But moles of HCl is equal to moles of
CaCO3, i.e., 0.1.
Thus, HCl is the limiting reagent
From reaction (i)
2 moles of HCl produce 1 mole of
CO2 Thus, 0.1 mole of HCl will produce =
×0.1 = 0.05 moles of
CO2 1 mole of
CO2 = 22.4 L of volume
= 22.4 ×
103cm3 (Since 1 L = 1000
cm3)
Thus, 0.05 moles of
CO2 = 0.05 × 22.4 ×
103 = 1120
cm3 Hence, option 'A' is correct