Question

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?

Answer 1

m CO₂ = 0.249 g

Step-by-step 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

`moles HCl = C_(HCl) \cdot V_(HCl) = 0.450 (mol)/(L) \cdot 0.025L = 0.0113 moles`

Now, we can find the moles of CO₂:

`moles CO_(2) = \frac {moles HCl}{2} = 5.65 \cdot 10^(-3) moles`

Finally, the mass of CO₂ is:

`m CO_(2) = moles _{CO_(2)} \cdot M_{CO_(2)} = 5.65 \cdot 10^(-3) moles \cdot 44.01 (g)/(mol) = 0.249 g CO_(2)`

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

Have a nice day!

Answer 2

the amount of CO2 gas produced is 0.260 gr

Step-by-step explanation:

Assuming that the antacid quantity is enough to react will all the HCl , the amount of HCl mass that reacts is:

n Hcl= M * V = 0.450 moles/L * 25 ml *1000 mmoles/moles *1 L/1000ml = 11.250 mmoles

According to the chemical equation , the amount of Co2 moles produced is the half of the number of Hcl moles that react

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

thus

n Co2 = n Hcl / 2 = 11.25 moles/2 = 5.625 moles

therefore , knowing that the molecular weight is Mw=44 gr/mol

m Co2 = n Co2 * Mw = 5.625 mmoles * 44 gr/mol * 1 mol/ 1000 mmoles = 0.260 gr