Problem 26 - Introductory Chemistry Atoms First

Question

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 \(\mathrm{ZnO}\) ?

Step-by-Step Solution

Answer

(a) Balanced equation: \[ 2 \mathrm{ZnS} + 3 \mathrm{O_{2}} \rightarrow 2 \mathrm{ZnO} + 2 \mathrm{SO_{2}} \] (b) Theoretical yields: - \(\mathrm{ZnO}: 16.93 \mathrm{\ g}\) - \(\mathrm{SO_{2}}: 13.34 \mathrm{\ g}\) (c) Excess reactant (\(\mathrm{ZnS}\)) left over: \(0.747 \mathrm{\ g}\) (d) Percent yield of \(\mathrm{ZnO}: 44.3 \%\).

Step 1: Write the unbalanced equation

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

Step 2: Balance the equation

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#

Step 1: Calculate the molar mass of the reactants and products

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 \(\mathrm{SO_{2}} = (32.07 + 2 \cdot 16) \mathrm{\ g/mol} = 64.07 \mathrm{\ g/mol}\)

Step 2: Calculate the moles of reactants

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 \(\mathrm{O_{2}} = \frac{10.0 \mathrm{\ g}}{32 \mathrm{\ g/mol}} = 0.3125 \mathrm{\ mol}\)

Step 3: Identify the limiting reactant

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.

Step 4: Calculate the theoretical yield of each product

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#

Step 1: Calculate the amount of \(\mathrm{ZnS}\) consumed

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 = \(\frac{2 \ moles \ \mathrm{ZnS}}{3 \ moles \ \mathrm{O_{2}}} \cdot 0.3125 \mathrm{\ mol} = 0.2083 \mathrm{\ mol}\)

Step 2: Calculate the amount of \(\mathrm{ZnS}\) left over in grams

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 \(\mathrm{ZnO}\)#

Step 1: Calculate the actual yield of \(\mathrm{ZnO}\)

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

Step 2: Calculate the percent yield of \(\mathrm{ZnO}\)

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

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is adhered to. This law states that matter is neither created nor destroyed in a chemical reaction. Thus, the number of atoms of each element in the reactants must be equal to the number of atoms in the products. The steps involved include writing the unbalanced equation and adjusting coefficients to balance the atoms on both sides without altering the chemical formulas.

Theoretical Yield Calculation

Theoretical yield calculation involves determining the maximum amount of product that can be produced in a chemical reaction under ideal conditions. It is based on the stoichiometry of the balanced equation and the limiting reactant—the reactant that will be completely consumed first, limiting the amount of product formed. The calculation begins with finding molar masses and converting the given masses of reactants into moles. Thereafter, one identifies the limiting reactant using mole ratios and calculates the moles of product expected. The final step is converting moles of product to grams to find the theoretical yield.

Limiting Reactant Determination

Identifying the limiting reactant is crucial because it dictates the maximum amount of product that can be formed in a chemical reaction. To determine the limiting reactant, one must calculate the mole ratio of the reactants and compare it to the mole ratio in the balanced chemical equation. The reactant that provides the lesser amount of product according to the stoichiometry is the limiting reactant. This approach helps to predict how much of the excess reactant will be leftover after the reaction has gone to completion.

Percent Yield Calculation

The percent yield is a measure of the efficiency of a chemical reaction and is calculated by comparing the actual yield (the amount actually obtained from the reaction) to the theoretical yield. The percent yield is determined using the formula:
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \% \].
The value indicates how close the experimental conditions were to ideal and can be impacted by factors such as incomplete reactions, side reactions, or loss of product during the process. Percent yield can exceed 100% if there are impurities present or if the actual yield is incorrectly measured.