100 g of caco3 is treated with 500ml of M/2 solution of HCL.Find out the volume of CO2 evolved at STP.
Number of moles of CaCO3 = 100 / 100 = 1.0
Moles HCl = 0.500 × M/2 = 0.25
so HCl IS THE LIMITING REAGENT HERE.
Following is the reaction..
CaCO3 + 2 HCl →CaCl2 + H2O + CO2
Moles of CO2 = 0.25/2=0.125
Volume of CO2 at STP = 0.125 × 22.4 =2.8 L