When powdered zinc is heated with sulfur, a violent reaction occurs, and zinc su

Question

When powdered zinc is heated with sulfur, a violent reaction occurs, and zinc sulfide forms: Zn(s) + S8(s) rightarrow ZnS(s) [unbalanced] When 85.2 g of Zn reacts with 52.4 g of S8, 105.4 g of ZnS forms. What is the percent yield of ZnS? Questions to consider: When you know amounts of all reactants in a reaction and must calculate an amount of a product, what must you find first that involves the stoichiometry of the reaction? What values do you need to calculate % yield? What part of calculating it will come from the stoichiometry of the reaction? Two additional side reactions also occur. The remaining Zn present reacts with oxygen in air to form zinc oxide, and the remaining sulfur reacts with oxygen to form sulfur dioxide. If all of the remaining Zn and Sg from the above gets consumed in these two reactions with oxygen, how many grams of each of the two oxides form? Questions to consider: How will you apply the % yield you calculated in (a) above to the stoichiometry of the reaction in order to find the number of moles of both Zn and S8 remaining after the reaction in (a)? What are the 2 separate reactions of each Zn and S8 with oxygen to form each one's respective oxides?

Explanation / Answer

Percent yield is 100(actual mass of product/ predicted mass of product)

100 * 105.4/ 97.445

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