Answer:
the amount of CO2 gas produced is 0.260 gr
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
Answer:
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
[tex] moles HCl = C_{HCl} \cdot V_{HCl} = 0.450 \frac{mol}{L} \cdot 0.025L = 0.0113 moles [/tex]
Now, we can find the moles of CO₂:
[tex] moles CO_{2} = \frac {moles HCl}{2} = 5.65 \cdot 10^{-3} moles [/tex]
Finally, the mass of CO₂ is:
[tex] m CO_{2} = moles _{CO_{2}} \cdot M_{CO_{2}} = 5.65 \cdot 10^{-3} moles \cdot 44.01 \frac{g}{mol} = 0.249 g CO_{2} [/tex]
Hence 0.249 g of CO₂ are produced from the reaction of calcium carbonate with HCl.
Have a nice day!
