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Chapter 9: Problem 164

Consider the unbalanced chemical equation \(\mathrm{As}_{4} \mathrm{~S}_{6}+\mathrm{O}_{2} \rightarrow \mathrm{As}_{4} \mathrm{O}_{6}+\mathrm{SO}_{2}\) (a) How many grams of \(\mathrm{O}_{2}\) are needed to react completely with \(58.9 \mathrm{~g}\) of \(\mathrm{As}_{4} \mathrm{~S}_{6}\) ? (b) If \(41.2 \mathrm{~g}\) of \(\mathrm{SO}_{2}\) is produced, what is the percent yield for the reaction?

Short Answer

Expert verified
(a) 28.26 grams of O2 are needed to react completely with 58.9 g of As4S6. (b) The percent yield for the reaction is approximately 54.55%.

Step by step solution

01

Balance the Chemical Equation

Balanced equation: 2As4S6 + 9O2 -> 2As4O6 + 12SO2 Now that we have the balanced equation, we can use stoichiometry to solve the given problems.
02

Calculate the Moles of As4S6

First, we need to find the moles of As4S6. To do this, we'll divide the given mass of As4S6 (58.9 g) by its molar mass. Molar mass of As4S6 = 4(74.92 g/mol for As) + 6(32.07 g/mol for S) = 299.68 g/mol Moles of As4S6 = (58.9 g)/(299.68 g/mol) = 0.1965 moles
03

Determine the Moles of O2 Required

Using the balanced equation and stoichiometry, we can find the moles of O2 required to react completely with As4S6. Mole ratio (O2 to As4S6) = 9:2 Moles of O2 = 0.1965 moles(As4S6) * (9 moles(O2) / 2 moles(As4S6)) = 0.8830 moles (O2)
04

Calculate the Mass of Oxygen Needed

To find the mass of O2, we will multiply the moles of O2 by its molar mass. Molar mass of O2 = 32.00 g/mol Mass of O2 = 0.8830 moles * 32.00 g/mol = 28.26 g So, 28.26 grams of O2 are needed to react completely with 58.9 g of As4S6. To answer (b), we need to calculate the percent yield of the reaction.
05

Calculate the Theoretical Moles of SO2 Produced

Using stoichiometry and the balanced equation, we can find the theoretical moles of SO2 produced. Mole ratio (SO2 to As4S6) = 12:2 Theoretical moles of SO2 = 0.1965 moles(As4S6) * (12 moles(SO2) / 2 moles(As4S6)) = 1.179 moles(SO2)
06

Calculate the Theoretical Mass of SO2 Produced

To find the theoretical mass of SO2, we will multiply the moles of SO2 by its molar mass. Molar mass of SO2 = 64.07 g/mol Theoretical mass of SO2 = 1.179 moles * 64.07 g/mol = 75.56 g
07

Calculate the Percent Yield of the Reaction

Now we can calculate the percent yield using the actual mass of SO2 produced (41.2 g) and the theoretical mass of SO2 (75.56 g). Percent yield = (Actual mass of SO2 / Theoretical mass of SO2) * 100% Percent yield = (41.2 g / 75.56 g) * 100% = 54.55% The percent yield for the reaction is approximately 54.55%.

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Most popular questions from this chapter

Aluminum metal burns in chlorine gas to form aluminum chloride. (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting if \(100.0 \mathrm{~g}\) of \(\mathrm{Al}\) and \(5.00\) moles of \(\mathrm{Cl}_{2}\) are used?

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