Calcium carbonate (caco3) reacts with stomach acid (hcl, hydrochloric acid) according to the following equation: caco3(s) 2hcl(aq)⟶co2(g) h2o(l) cacl2(aq) a typical antacid contains caco3. if such an antacid is added to 25.0 ml of a solution that is 0.450 m in hcl, how many grams of co2 gas are produced?

m CO₂ = 0.249 g

Explanation:

Assuming that CaCO₃ is in excess, from the next reaction we can calculate the moles of CO₂ from the moles of HCl:      

CaCO₃(s) + 2HCl(aq) → CO₂(g) + H₂O(l) + CaCl₂(aq)

                V: 25.0mL     m=?

                C: 0.450M     M: 44.01g/mol

Now, we can find the moles of CO₂:

Finally, the mass of CO₂ is:  

Hence 0.249 g of CO₂ are produced from the reaction of calcium carbonate with HCl.    

Have a nice day!

a:

by the combination of an unlimited number of elements