The balanced chemical equation for the reaction between hydrochloric acid (HCl) and calcium carbonate (CaCO₃) is:
CaCO₃+2HCl→CaCl₂+H₂O+CO₂
From the equation, we see that 1 mole of CaCO₃ reacts with 2 moles of HCl to produce 1 mole of CO₂.
First, we calculate the number of moles of each reactant:
Moles of HCl:
Molar mass of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol
Moles of HCl = (1.0 g)/(36.5 g/mol)≈0.027 moles
Moles of CaCO₃:
Molar mass of CaCO₃ = 40.1 (Ca) + 12.0 (C) + 3 x 16.0 (O) = 100.1 g/mol
Moles of CaCO₃ =(1.0 g)/(100.1 g/mol)≈0.01 moles
Next, determine the limiting reactant. According to the stoichiometry:
1 mole of CaCO₃ requires 2 moles of HCl.
0.01 moles of CaCO₃ would require 0.01×2=0.02 moles of HCl.
Since we have only 0.027 moles of HCl, which is more than the required 0.02 moles, CaCO₃ is the limiting reactant.
Since CaCO₃ is the limiting reactant, the amount of CO₂ produced is based on the moles of CaCO₃.
1 mole of CaCO₃ produces 1 mole of CO₂.
0.01 moles of CaCO₃ will produce 0.01 moles of CO₂.
Finally, calculate the volume of CO₂ produced at NTP (Normal Temperature and Pressure), where 1 mole of any gas occupies 22.4 liters.
Volume of CO₂ = 0.01 moles × 22.4 liters/mole = 0.224 liters.
Hence, the correct answer is option b. 0.224 liters
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and calcium carbonate (CaCO₃) is:
CaCO₃+2HCl→CaCl₂+H₂O+CO₂
From the equation, we see that 1 mole of CaCO₃ reacts with 2 moles of HCl to produce 1 mole of CO₂.
First, we calculate the number of moles of each reactant:
Moles of HCl:
Molar mass of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol
Moles of HCl = (1.0 g)/(36.5 g/mol)≈0.027 moles
Moles of CaCO₃:
Molar mass of CaCO₃ = 40.1 (Ca) + 12.0 (C) + 3 x 16.0 (O) = 100.1 g/mol
Moles of CaCO₃ =(1.0 g)/(100.1 g/mol)≈0.01 moles
Next, determine the limiting reactant. According to the stoichiometry:
1 mole of CaCO₃ requires 2 moles of HCl.
0.01 moles of CaCO₃ would require 0.01×2=0.02 moles of HCl.
Since we have only 0.027 moles of HCl, which is more than the required 0.02 moles, CaCO₃ is the limiting reactant.
Since CaCO₃ is the limiting reactant, the amount of CO₂ produced is based on the moles of CaCO₃.
1 mole of CaCO₃ produces 1 mole of CO₂.
0.01 moles of CaCO₃ will produce 0.01 moles of CO₂.
Finally, calculate the volume of CO₂ produced at NTP (Normal Temperature and Pressure), where 1 mole of any gas occupies 22.4 liters.
Volume of CO₂ = 0.01 moles × 22.4 liters/mole = 0.224 liters.
Hence, the correct answer is option b. 0.224 liters