What volume of CO2 (g) measured at STP, can be obtained by the reaction of 68.0 g CaCo3 with excess hydrochloric acid? CaCO3 (s) + 2HCl (aq) -> CaCl2 (aq) + CO2 (g) + H2O
`CaCO_3+2HCl rarr CaCl_2+CO_2+H_2O`
`CaCO_3:CO_2 = 1:1`
Molar mass of `CaCO_3 = 100g/(mol)`
Amount of `CaCO_3` reacted `= 68/100 = 0.68mol`
Amount of `CO_2` formed `= 0.68mol`
At STP volume of 1mol of gas substance is `22.4l` .
Therefore volume of 0.68mol `CO_2 = 22.4xx0.68 = 15.232l`
So the answer is `15.232l`
`CO_2` act as an ideal gas.