Chemical treatment of zinc sulfide, \(\mathrm{ZnS}\), with oxygen, \(\mathrm{O}_{2}\), gives zinc oxide, \(\mathrm{ZnO}\), and sulfur dioxide gas, \(\mathrm{SO}_{2}\). (a) Write a balanced equation for the reaction. (b) If \(10.0 \mathrm{~g}\) of \(\mathrm{Zn} \mathrm{S}\) is combined with \(10.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\), what is the theoretical yield of each product, in grams? (c) How much of the excess reactant is left over, in grams? (d) Suppose only \(7.50 \mathrm{~g}\) of \(\mathrm{Zn} \mathrm{O}\) is recovered. What is the percent yield of \(\mathrm{ZnO}\) ?

The unbalanced equation for the reaction is: \[ \mathrm{ZnS} + \mathrm{O_{2}} \rightarrow \mathrm{ZnO} + \mathrm{SO_{2}} \]

To balance the equation, we need the equal number of atoms of each element on both sides of the equation. \[ 2 \mathrm{ZnS} + 3 \mathrm{O_{2}} \rightarrow 2 \mathrm{ZnO} + 2 \mathrm{SO_{2}} \] Now the equation is balanced. #b) Determining the theoretical yield#

We need to find the molar mass of \(\mathrm{ZnS}\), \(\mathrm{O_2}\), \(\mathrm{ZnO}\), and \(\mathrm{SO_{2}}\). Molar mass of \(\mathrm{ZnS} = (65.38 + 32.07) \mathrm{\ g/mol} = 97.45 \mathrm{\ g/mol}\) Molar mass of \(\mathrm{O_{2}} = 2 \cdot (16) \mathrm{\ g/mol} = 32 \mathrm{\ g/mol}\) Molar mass of \(\mathrm{ZnO} = (65.38 + 16) \mathrm{\ g/mol} = 81.38 \mathrm{\ g/mol}\) Molar mass of \(\mathrm{SO_{2}} = (32.07 + 2 \cdot 16) \mathrm{\ g/mol} = 64.07 \mathrm{\ g/mol}\)

Given mass of \(\mathrm{ZnS} = 10.0 \mathrm{\ g}\) Given mass of \(\mathrm{O_{2}} = 10.0 \mathrm{\ g}\) Moles of \(\mathrm{ZnS} = \frac{10.0 \mathrm{\ g}}{97.45 \mathrm{\ g/mol}} = 0.1027 \mathrm{\ mol}\) Moles of \(\mathrm{O_{2}} = \frac{10.0 \mathrm{\ g}}{32 \mathrm{\ g/mol}} = 0.3125 \mathrm{\ mol}\)

From the balanced equation, the mole ratio of \(\mathrm{ZnS}\) to \(\mathrm{O_{2}}\) is \(2 : 3\). We will compare the mole ratios of the reactants to find which one is limiting. \(\frac{0.1027 \mathrm{\ mol}\ \mathrm{ZnS} }{0.3125 \mathrm{\ mol} \ \mathrm{O_{2}}} = \frac{x}{3}\) Calculate the value of \(x\): \(x = \frac{0.1027 \mathrm{\ mol} \ \mathrm{ZnS}}{0.3125 \mathrm{\ mol}\ \mathrm{O_{2}}} \cdot 3 = 0.987 \mathrm{\ mol}\) Since \(0.987 \mathrm{\ mol} > 2 \mathrm{\ mol}\), it means that \(\mathrm{O_{2}}\) is the limiting reactant.

To find the theoretical yield of each product, we will use the limiting reactant, \(\mathrm{O_{2}}\), and apply the mole ratios from the balanced equation. Moles of \(\mathrm{ZnO} = \frac{2 \ moles \ \mathrm{ZnO}}{3 \ moles \ \mathrm{O_{2}}} \cdot 0.3125 \ moles = 0.2083 \mathrm{\ mol}\) Moles of \(\mathrm{SO_{2}} = \frac{2 \ moles\ \mathrm{SO_{2}}}{3 \ moles \ \mathrm{O_{2}}} \cdot 0.3125 \ moles = 0.2083 \mathrm{\ mol}\) Now convert moles of each product to grams. Theoretical yield of \(\mathrm{ZnO} = 0.2083 \mathrm{\ mol} \cdot 81.38 \mathrm{\ g/mol} = 16.93 \mathrm{\ g}\) Theoretical yield of \(\mathrm{SO_{2}} = 0.2083 \mathrm{\ mol} \cdot 64.07 \mathrm{\ g/mol} = 13.34 \mathrm{\ g}\) #c) Amount of excess reactant left over#

Using the mole ratio from the balanced equation, calculate the moles of \(\mathrm{ZnS}\) consumed by combining with the \(\mathrm{O_{2}}\). Moles of \(\mathrm{ZnS}\) consumed = \(\frac{2 \ moles \ \mathrm{ZnS}}{3 \ moles \ \mathrm{O_{2}}} \cdot 0.3125 \mathrm{\ mol} = 0.2083 \mathrm{\ mol}\)

Subtract the moles of \(\mathrm{ZnS}\) consumed from the initial moles of \(\mathrm{ZnS}\). Moles of \(\mathrm{ZnS}\) left = \(0.1027 \mathrm{\ mol} - 0.2083 \mathrm{\ mol} = 0.00767 \mathrm{\ mol}\) Now convert the moles of left over \(\mathrm{ZnS}\) to grams. Amount of \(\mathrm{ZnS}\) left over = \(0.00767 \mathrm{\ mol} \cdot 97.45 \mathrm{\ g/mol} = 0.747 \mathrm{\ g}\) #d) Percent yield of \(\mathrm{ZnO}\)#

According to the problem, the actual yield of \(\mathrm{ZnO}\) is \(7.50 \mathrm{\ g}\).

Percent yield is calculated by dividing the actual yield by the theoretical yield and then multiplying by \(100\). Percent yield of \(\mathrm{ZnO} = \frac{7.50 \mathrm{\ g}}{16.93 \mathrm{\ g}} \cdot 100 = 44.3 \%\)