Question

Zinc reacts with sulfur according to the equation Zn(s) + S(s) → ZnS(s) In an experiment, 6.54 g of zinc was reacted with sulfur to make zinc sulfide, ZnS. 9.00 g of zinc sulfide was obtained. Calculate the percentage yield of zinc sulfide.​

Answer 1

The percentage yield of zinc sulfide is 86.0%.

Step-by-step explanation:

In the given reaction, zinc (Zn) reacts with sulfur (S) to form zinc sulfide (ZnS) according to the balanced chemical equation:
`\(Zn(s) + S(s) \rightarrow ZnS(s)\)`. The experimental yield of zinc sulfide is the actual amount obtained in the experiment, which is 9.00 g. The theoretical yield is the maximum amount of product that could be obtained based on the stoichiometry of the reaction.

To calculate the theoretical yield, we first need to determine the molar mass of zinc (Zn) and sulfur (S) from the periodic table. The molar mass of Zn is approximately 65.38 g/mol, and S is approximately 32.06 g/mol. The balanced equation shows a 1:1 ratio between Zn and ZnS, so the molar mass of ZnS is the sum of the molar masses of Zn and S:
`\(65.38 + 32.06 = 97.44\) g/mol.`

Now, we can use the molar masses to find the theoretical yield of ZnS from 6.54 g of Zn:

`\[\text{Theoretical yield} = \frac{\text{Mass of Zn}}{\text{Molar mass of Zn}} * \text{Molar mass of ZnS} = \frac{6.54 \, \text{g}}{65.38 \, \text{g/mol}} * 97.44 \, \text{g/mol} \approx 9.70 \, \text{g}\]`

The percentage yield is then calculated using the formula:

`\[\text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) * 100 = \left( \frac{9.00 \, \text{g}}{9.70 \, \text{g}} \right) * 100 \approx 86.0\%\]`

Therefore, the percentage yield of zinc sulfide is 86.0%, indicating that 86.0% of the theoretically calculated amount was obtained in the experiment. Factors such as incomplete reactions, side reactions, or experimental errors may contribute to a percentage yield less than 100%.