Question

When powdered zinc is heated with sulfur, a violent reaction occurs, and zinc sulfide forms:

$\mathit{Z}\mathit{n}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathit{S}}_{\mathbf{8}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathit{Z}\mathit{n}\mathit{S}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\left[}\mathit{u}\mathit{n}\mathit{b}\mathit{a}\mathit{l}\mathit{a}\mathit{n}\mathit{c}\mathit{e}\mathit{d}\mathbf{\right]}$

Some of the reactants also combine with oxygen in air to form zinc oxide and sulfur dioxide. When 83.2 g of Zn reacts with 52.4 g of S₈, 104.4 g of ZnS forms. What is the percent yield of ZnS? (b) If all the remaining reactants combine with oxygen, how many grams of each of the two oxides form?

Short answer

1. The percent yield is84.3%.
2. There are 16.4 g ZnO and 35.8 g SO₂ produced

Step-by-step solution

Finding the percent yield

The moles of ZnS can be found as,

$$\begin{gathered} MolesofZnS\left(fromZn\right)=83.2gZn\times \frac{1molZn}{65.38gZn}\times \frac{8molZnS}{8molZn}=1.27molZnS \\ MolesofZnS\left(from{S}_8\right)=52.4g{S}_8\times \frac{1mol{S}_8}{256.56g{S}_8}\times \frac{8molZnS}{1mol{S}_8}=1.63molZnS \end{gathered}$$

Calculate mass of zinc and its percent yield:

The mass of Zn and percent yield can be denoted as,

$$\begin{gathered} MassofZn\left(theoretical\right)=1.27molZnS\times \frac{97.47gZnS}{1molZnS}=123.8gZnS \\ \%yield=\frac{actualyield}{theoreticalyield}\times 100 \\ \%yield=\frac{104.4g}{123.8g}\times 100 \\ \%yield=84.3\% \end{gathered}$$

Finding the production of each of the two oxides

The mass for Zn and S₈ can be formed as,

$$\begin{gathered} MassofZn(forZnS)=104.4gZnS\times \frac{1molZnS}{97.47gZnS}\times \frac{8molZn}{8molZnS}\times \frac{65.38gZn}{1molZn}=70gZn \\ Massof{S}_8(forZnS)=104.4gZnS\times \frac{1molZnS}{97.47gZnS}\times \frac{1mol{S}_8}{8molZnS}\times \frac{256.56g{S}_8}{1mol{S}_8}=34.5g{S}_8 \end{gathered}$$

On subtracting these masses,

$$\begin{gathered} MassofZn(fromZnO)=83.2g-70g=13.2gZn \\ Massof{S}_8(forS{O}_2)=52.4g-34.5g=17.9S_8 \end{gathered}$$

The mass of produced oxides as follows,

$$\begin{gathered} MassofZnO=13.2gZn\times \frac{1molZn}{65.38gZn}\times \frac{2molZnO}{2molZn}\times \frac{81.38gZnO}{1molZnO}=16.4gZnO \\ MassofS{O}_2=17.9g{S}_8\times \frac{1mol{S}_8}{256.56g{S}_8}\times \frac{8molSO}{1mol{S}_8}\times \frac{64.06gS{O}_2}{1molS{O}_2}=35.8gS{O}_2 \end{gathered}$$