Question

Sulfur dioxide gas reacts with sodium hydroxide to form sodium sulfite and water. Theunbalanced chemical equation for this reaction is given below:

$\mathit{S}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathit{N}\mathit{a}\mathit{O}\mathit{H}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathit{N}{\mathit{a}}_{\mathbf{2}}\mathit{S}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}$

Assuming you react 38.3 g sulfur dioxide with 32.8 g sodium hydroxide and assuming that the reaction goes to completion, calculate the mass of each product formed.

Short answer

The mass of each product formed, Na₂SO₃ is 51.7g and H₂O is 7.39g.

Step-by-step solution

Definition

The molar mass of a material is defined as the mass of the substance contained in one mole of the substance, and it is equal to the total mass of the component atoms. It may be stated mathematically as follows:

Molar mass$=\frac{Weighting}{numberofmoles}$

Determination of molar mass of SO2 and NaOH

First and foremost, let’s balance the equation

$S{O}_2\left(g\right)+2NaOH\left(s\right)\to N{a}_{2}S{O}_3\left(s\right)+H_{2}O\left(l\right)$

Molar mass of SO₂ :

$32.07\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.+16.00\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\times 64.07\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Molar massof NaOH :

$22.99\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.+16.00\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.+1.01\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.=40.00\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Determination of molar mass of Na2SO3 and water

Molar mass of Na₂SO₃:

$22.99\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\times 2+32.07\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.+16.00\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\times 3=126.05\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Molar mass ofwater:

$1.01\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\times 2+16.00\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.=18.02\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Determination of limiting reactant

We need to figure out which reaction is the limiting one.

This can be accomplished bychanging a mole of one reactant to a mole of another reactant:

Moles of NaOH :

$32.8g\times \frac{1mol}{40.00g}=0.82mol$

Let’s convert SO₂ to NaOH

$32.8g\times \frac{1molS{O}_2}{64.07g}\times \frac{2molNaOH}{1molS{O}_2}=1.20mol$

As, 0.82 mol<1.20 mol,

The limiting reactant is NaOH.

Determination of mass of products

Let’s determine the mass of products,

$$\begin{gathered} 0.82\mathrm{mol}\mathrm{NaOH}\times \frac{1\mathrm{mol}{\mathrm{Na}}_2{\mathrm{SO}}_3}{2\mathrm{mol}\mathrm{NaOH}}\times \frac{126.05\mathrm{g}{\mathrm{Na}}_2{\mathrm{SO}}_3}{1\mathrm{mol}}=51.7\mathrm{g}{\mathrm{Na}}_2{\mathrm{SO}}_3 \\ 0.82\mathrm{mol}\mathrm{NaOH}\times \frac{1\mathrm{mol}{\mathrm{H}}_2\mathrm{O}}{2\mathrm{mol}\mathrm{NaOH}}\times \frac{18.02\mathrm{g}{\mathrm{H}}_2\mathrm{O}}{1\mathrm{mol}}=7.39\mathrm{g}{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Therefore the mass of Na₂SO₃is 51.7gand H₂Ois 7.39g .