Question

Acodofocation of sodium bicarbonate(NaHCO₃) produces carbon dioxide in a two-step process. The first step produces carbonic acid (H₂CO₃). Write and balance the molecular equation for the first step of the reaction of sodium bicarbonate with HF.

The second step is the decompostion of a carbonic acid to ccarbon dioxide. Write and balance the molecular equation for the second step.

Now write the overall net ionic equation for te acidification of NaHCO₃ to produce CO₂.

What is the molar ratio of CO₂ produced to NaHCO₃ decomposed?

What mass of CO₂ will be produced by the complete decompostion of 1346 mg of NaHCO₃?

Answer 1

Answer –

The balance molecular equation for the first step of the reaction of sodium bicarbonate with HF –

     NaHCO₃(aq)+ HF(aq) ------> H₂CO₃(aq)+ NaF(aq)

The balance molecular equation for the second step –

H₂CO₃(aq) <-----> CO₂(g)+ H₂O(l)

The overall net ionic reaction –

Total ionic reaction –

Na⁺(aq) + HCO₃^-(aq) + H⁺(aq) + F^-(aq) ------> Na⁺(aq) + F^-(aq) + CO₂(g) + H₂O(l)

Net ionic reaction -

HCO₃^-(aq) + H⁺(aq) -----> CO₂(g) + H₂O(l)

Given, mass of NaHCO₃(aq) = 1346 mg

We need to convert it in to mg to g

1 mg = 0.001 g

1346 mg = ?

= 1.346 g

Moles of NaHCO₃ = 1.346 g / 84.006 g.mol^-1

                             = 0.0160 moles

From the above balanced equation –

1 moles of NaHCO₃ = 1 moles of CO₂

So, 0.0160 moles of NaHCO₃ = ?

= 0.0160 moles CO₂

Mass of CO₂ = 0.0160 moles * 44.009 g/mol

                      = 0.705 g

                      = 705 mg

705 mg of mass of CO₂ will be produced by the complete decomposition of 1346 mg of NaHCO₃