Sodium bicarbonate is reacted with concentrated hydrochloric acid at 37.0 °c and 1.00 atm. the reaction of 6.00 kg of bicarbonate with excess hydrochloric acid under these conditions will produce l of co2.

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

the answers would be

the first one and the second one