Question

The volume of CO2 liberated at STP if 50 g of CaCO3 is treated with dil.Hcl acid which contains 7.3 g of dissolved Hcl gas.

Answer 1

CaCO3+ 2HCl--------> CaCl2+H2O+CO2

From the Q: we have 0.5 moles (50/100) of CaCO3 & 0.2 moles (7.3/36.5) of HCl

Since HCl seems to be the LIMITING REAGENT so we do the problem using it only.

Acc to the eqn : 2 moles HCl-------------->1 mole of CO2

                          0.2 moles------------------> ?

                 =0.1 mole CO2 *22.4L

                 = 2.24 L of CO2 is liberated at STP.

Answer 2

Answer:

You need to start off by writing out a balanced chemical equation which will show you the relationship between CaCO3 and CO2:

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

This equation tells you that for every mol of CaCO3 consumed you will be producing 1 mol of CO2. In other words there is a 1:1 relationship between CaCO3 and CO2.

The next step is to determine the amount of CaCO3 you have in moles, and since HCl is in excess, you do not need to worry about which one is the limiting reagent (CaCO3 is limiting since HCl is in excess). You can do that by using the molar mass of CaCO3, which is 100 g/mol:

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