Question

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.

Answer 1

Answer:

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

Answer 2

Answer:

1837.89 Lt

Explanation:

The chemical reaction for this situation is:

NaHCO₃ + HCl → NaCL + H₂O + CO₂ ₍g₎

Where the mola mass we need are:

M NaHCO₃ = 84 g/mol

M CO₂ = 44 g/mol

As we have 6.00 Kg of sodium bicarbonate, then:

6 Kg NaHCO₃ = 71.43 moles of NaHCO₃

Due the stoichiometry of this chemaicl reaction:

1 mol NaHCO₃ = 1 mol CO₂

71.43 moles NaHCO₃ = 71.43 moles CO₂

And considering that CO₂ is an ideal gas, we can use the following formula:

PV=nRT

V = (nRT)/P

n = 71.43 mol

R = 0.083 Ltxatm(molxK)

T = 37°C = 310 K

P = 1 atm

So: V = (71.43x0.083x310)/1

V CO₂ = 1837.89 Lt