Question: 6.3 g of NaHCO₃ reacts with 10 g of HCl solution to form NaCl and CO₂. If 12 g of NaCl is formed, what is the mass of CO₂ gas released in the reaction?
Answer: (c) 2.8 g ✅
Explanation:
The reaction is:
NaHCO₃+HCl→NaCl+H₂O+CO₂\text{NaHCO₃} + \text{HCl} → \text{NaCl} + \text{H₂O} + \text{CO₂}
- Mass of reactants = Mass of products (Law of Conservation of Mass)
Let mCO₂m_{\text{CO₂}} be the mass of CO₂:
Mass of NaHCO₃ + Mass of HCl=Mass of NaCl + Mass of CO₂ + Mass of H₂O\text{Mass of NaHCO₃ + Mass of HCl} = \text{Mass of NaCl + Mass of CO₂ + Mass of H₂O}
Assuming H₂O is not mentioned, the problem simplifies to:
6.3+10=12+mCO₂6.3 + 10 = 12 + m_{\text{CO₂}} mCO₂=16.3−12=4.3 g?m_{\text{CO₂}} = 16.3 - 12 = 4.3 \text{ g?}
Wait, the options don’t match exactly. Let’s calculate using stoichiometry instead:
Molar masses:
- NaHCO₃ = 84 g/mol
- NaCl = 58.5 g/mol
- CO₂ = 44 g/mol
From reaction: 1 mole NaHCO₃ → 1 mole NaCl + 1 mole CO₂
Mass ratio of NaCl to CO₂: 58.5 : 44
Let x = mass of CO₂ formed for 12 g NaCl:
58.544=12x ⟹ x=12×4458.5≈9.02 g\frac{58.5}{44} = \frac{12}{x} \implies x = \frac{12 \times 44}{58.5} \approx 9.02 \text{ g}
Correct Answer: (d) 9.1 g ✅
So the mass of CO₂ released is approximately 9.1 g.
Do you want me to explain why my first shortcut method gave a wrong result?