You carefully weigh out 20.00g of \rm CaCO_3 powder and add it to 81.00g of \rm HCl solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 92.80g . The relevant equation is

\rm CaCO_3{\it (s)}+2HCl{\it (aq)}\rightarrow H_2O{\it (l)}+CO_2{\it (g)}+CaCl_2{\it (aq)}
Assuming no other reactions take place, what mass of \rm CO_2 was produced in this reaction?

Something about this problem doesn't add up.

If you go with 20 g CaCO3, that is 20/100 = 0.2 mol and that will produce 0.2 x 44 = 8.80g CO2.

However, if you add 81.00 g to 20 g to make 101 g total and you reweigh after the CO2 is expelled, the loss in mass is 101.0-92.80 = 8.20g.
It seems to me that the answers should be the same no matter which way the problem is worked.

To find the mass of CO2 produced in this reaction, we need to use stoichiometry and the concept of the mole.

First, let's find the number of moles of CaCO3 and HCl using their given masses and molar masses. The molar mass of CaCO3 is 100.09 g/mol, and the molar mass of HCl is 36.45 g/mol.

The number of moles of CaCO3 can be calculated using the formula:

moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3

moles of CaCO3 = 20.00 g / 100.09 g/mol
moles of CaCO3 = 0.1999 mol (rounded to four decimal places)

Similarly, the number of moles of HCl can be calculated:

moles of HCl = mass of HCl / molar mass of HCl

moles of HCl = 81.00 g / 36.45 g/mol
moles of HCl = 2.2219 mol (rounded to four decimal places)

From the balanced equation, we can see that the stoichiometric ratio between CaCO3 and CO2 is 1:1. This means that for every one mole of CaCO3 reacted, one mole of CO2 is produced.

Therefore, the number of moles of CO2 produced is equal to the number of moles of CaCO3:

moles of CO2 = moles of CaCO3 = 0.1999 mol

Now, we need to calculate the mass of CO2. To do this, we'll use the formula:

mass of CO2 = moles of CO2 x molar mass of CO2

The molar mass of CO2 is 44.01 g/mol.

mass of CO2 = 0.1999 mol x 44.01 g/mol
mass of CO2 = 8.798 g (rounded to three decimal places)

Therefore, the mass of CO2 produced in this reaction is approximately 8.798 grams.