Question

100 g of CaCO3 is treated with 500 mL M/2 solution of HCl. Find out the volume of CO2 evolved at
S.T.P. Which substance is limiting reagent ?​

Answer 1

100g caco3=1 mole 500ml m/2 hcl= 1/4 mole as the stochiometric equation 1mol caco3=2 mol hcl=1 mol CO2 hence hcl is the limiting reagent since 2 mol hcl produce 1mol CO2 1/4 mol would produce =1/8 mol CO2=22.4/8=2.8 litre

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