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


M
oles 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

 

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