When 10 g of 90% pure limestone is heated, the approximate volume (in L) of `CO_2` liberated at STP is

Options

A. 4.4
B. `2.0`
C. `4.0`
D. 22.4

Answer

B

Explanation

Limestone `(CaCO_(3))` molar mass = 100 g/mol 10 g of 100% pure limestone = 10 g of `CaCO_(3)` But 10 g of 90% pure limestone
= 9 g of `CaCO_(3)`
9 g of `CaCO_(3)=0.09` mole of `CaCO_(3)`
Chemical reaction involved on heating `CaCO_(3)` is given by the equation.
`underset(1" mol")(CaCO_(3))overset(Delta)toCaO+underset(1" mol")(CO_(2)(g))`
Thus, 1 mole of `CaCO_(3)` yields 1 mole of `CO_(2)(g)` or 22.4 L of `CO_(2)(g)`.
Hence, 0.09 mole of `CaCO_(3)` will yield
`=0.09xx22.4" L of CO"_(2)`
`=2.016" L of "CO_(2)`.