Question

What volume of carbon dioxide gas can be collected from

thereaction between 10.0 g of calcium carbonate and 100.0 mL of
0.50 Mhydrochloric acid at 25.0oC and 1.0 atm
pressure?

Answer 1

1.22 L of carbon dioxide gas

Step-by-step explanation:

The reaction that takes place is:

• CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

First we determine which reactant is limiting:

• Calcium carbonate ⇒ 10.0 g CaCO₃ ÷ 100 g/mol = 0.10 mol CaCO₃

• Hydrochloric acid ⇒ 0.100 L * 0.50 M = 0.05 mol HCl

So HCl is the limiting reactant.

Now we calculate the moles of CO₂ produced:

• 0.05 mol HCl *
  `(1molCO_(2))/(1molHCl)` = 0.05 mol CO₂

Finally we use PV=nRT to calculate the volume:

• P = 1 atm

• n = 0.05 mol

• T = 25 °C ⇒ 25 + 273.16 = 298.16 K

1 atm * V = 0.05 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

• V = 1.22 L