Question

3. (a) Calcium carbonate reacts with hydrochloric acid to produce carbon dioxide gas. If 35.3 g of calcium carbonate reacts with 100 mL of 6.00 M HCI, how many liters of carbon dioxide gas will be produced at 745 mmHg and 23.0°C?

Answer 1

Q3. (a.) The balanced reaction is : CaCO₃ (s) + 2 HCl (aq) $\rightarrow$ CaCl₂ (aq) + CO₂ (g) + H₂O (l)

Case 1 : CaCO₃ is the limiting reactant

mass CaCO₃ = 35.3 g

moles CaCO₃ = (mass CaCO₃) / (molar mass CaCO₃)

moles CaCO₃ = (35.3 g) / (100.09 g/mol)

moles CaCO₃ = 0.3527 mol

moles CO₂ produced = moles CaCO₃ consumed

moles CO₂ produced = 0.3527 mol

Case 2 : HCl is the limiting reactant

concentration HCl = 6.00 M

volume HCl = 100 mL = 0.100 L

moles HCl = (concentration HCl) * (volume HCl in Liter)

moles HCl = (6.00 M) * (0.100 L)

moles HCl = 0.600 mol

moles CO₂ produced = (moles HCl) * (1 mole CO₂ / 2 moles HCl)

moles CO₂ produced = (0.600 mol) * (1/2)

moles CO₂ produced = 0.300 mol

Since less moles of CO₂ are produced in Case 2, therefore, HCl is the limiting reagent.

moles CO₂ produced = 0.300 mol

According to ideal gas law,

volume CO₂ produced = [(moles CO₂ produced) * (R) * (temperature)] / (pressure)

where

R = 0.0821 L-atm/mol-K

temperature = 23.0 ^oC = (23.0 + 273) K = 296 K

pressure = 745 mmHg

pressure = 745 mmHg * (1 atm / 760 atm)

pressure = 0.980 atm

Substituting the values,

volume CO₂ produced = [(0.300 mol) * (0.0821 L-atm/mol-K) * (296 K)] / (0.980 atm)

volume CO₂ produced = 7.44 L