posted by Dana on .
Calcium carbonate CaCO3 reacts with stomach acid (HCl, hydrochloric acid) according to the following equation:
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?
Here is a sample problem I've posted on simply stoichiometry. Just follow the steps. moles in this case = M x L.
do i still need to find whether CaCO3 or HCl is the limiting reactant?
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.