An antacid tablet weighing 0.853g contained calcium carbonate as the active ingredient, in addition to an inert binder. when an acid solution weighing 56.519g was added to the tablet, carbon dioxide gas was released, producing a fizz. the resulting solution weighed 57.152g. how many grams of carbon dioxide were produced?

0.22 g of CO2 were produced.

Explanation:

First, let's represent what is happening with an hypothetical chemical equation just to have a clearer vision of the presented process:

CaCO3 (aq) + 2 HAc (aq) → CaAc2 (aq) + H2O (l] + CO2 (g)

We have a tablet that has CaCO3 as the active ingredient that when combined with an acid, in this case represented as HAc, reacts producing a Calcium salt, water and carbon dioxide that will leave the solution as gas.

Having said that, we know that the initial mass of the reactants will have to maintain during the chemical reaction, or what is the same, the quantity of matter during the process will not change. So, if we have a tablet that weighs 0.853 g and we add an acid solution of 56.519 g, then we have that the initial mass of the reactants will be:

0.853 g from tablet + 56.519 g from acid solution = 57.372 g

This amount of matter should be the same after the reaction, but we know that the CO2 gas will leave the solution once it's formed, so considering that the resulting solution weighs 57.152 g we could calculate the mass of CO2 produced:

57.372 g of initial mass - 57.152 g of resulting solution = 0.22 g of CO2 that left the aqueous solution as gas.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole