Calcium carbonate CaCO3 reacts with stomach acid (HCl, hydrochloric acid) according to the following equation:

CaCO3(s)+2HCl(aq)->CaCl2(aq)+H2O(l)+CO2(g)
Tums, an antacid, contains CaCO3.
If Tums is added to 35.0 mL of 0.300 M HCl,how many grams of CO2 gas are produced?

To be a limiting reagent type problem BOTH reactants must be given; i.e., you have 35.0 mL of 0.300 M HCl (one of the reactants) but you don't have the other one (only that it is TUMS but no mass is given nor a percent composition--I thing the assumption is that you have enough CCO3 to use all of the HCl).

moles HCl = 0.035 x 0.300 = 0.0105

Convert moles HCl to moles CO2 using the coefficients in the balanced equation.
0.0105 moles HCl x (1 mole CO2/2 moles HCl) = 0.0105 x (1/2) = 0.00525
Now convert moles CO2 to grams.
g = moles x molar mass = ?? grams CO2.

Here is a sample problem I've posted on simply stoichiometry. Just follow the steps. moles in this case = M x L.

http://www.jiskha.com/science/chemistry/stoichiometry.html

do i still need to find whether CaCO3 or HCl is the limiting reactant?

thanks...

what volume of 0.150 m sotomach acid will caco3 nutralise ?

To find the number of grams of CO2 gas produced, we need to use stoichiometry and molar mass.

First, let's determine the number of moles of HCl in the solution:
moles of HCl = Molarity × volume
moles of HCl = 0.300 mol/L × 0.0350 L
moles of HCl = 0.0105 mol

According to the balanced equation, 1 mole of CaCO3 produces 1 mole of CO2 gas. Therefore, the number of moles of CO2 gas produced is also 0.0105 mol.

Now, we can calculate the mass of CO2 gas produced using its molar mass. The molar mass of CO2 is:
molar mass of CO2 = 12.01 g/mol (molar mass of C) + 2(16.00 g/mol) (molar mass of O)
molar mass of CO2 = 44.01 g/mol

Finally, we can calculate the mass of CO2 gas produced:
mass of CO2 gas = number of moles × molar mass
mass of CO2 gas = 0.0105 mol × 44.01 g/mol
mass of CO2 gas = 0.462 g

Therefore, approximately 0.462 grams of CO2 gas are produced when Tums is added to 35.0 mL of 0.300 M HCl.