1.82 × 10³ L
Explanation:
Let's consider the following reaction.
NaHCO₃ + HCl → NaCl + CO₂ + H₂O
The molar mass of NaHCO₃ is 84.01 g/mol. The moles corresponding to 6.00 kg (6.00 × 10³ g) of NaHCO₃ are:
6.00 × 10³ g × (1 mol/84.01 g) = 71.4 mol
The molar ratio of NaHCO₃ to CO₂ is 1:1. The moles of CO₂ are 71.4 moles.
We can find the volume of CO₂ using the ideal gas equation.
P × V = n × R × T
1.00 atm × V = 71.4 mol × (0.08206 atm.L/mol.K) × 310.2 K
V = 1.82 × 10³ L