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Chapter 7 Yield Calculations <Chapter 7 Yield Calculations Simulation Limiting Reagent and Theoretical Yield with Masses 7 of 13 You are provided with 5.80 g of S02 to react with 2.45 g of O. Indicate your theoretical yield by completing following statemonts Match the chemical formulas and amounts in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. View Available Hint(s) Reset Heip O2 The limiting reagent in the reaction is 0.0905 mol of SO, can Regarding the theoretical yield based on the limiting reagent 0.0766 be formed, which corresponds to a mass of 1.00 , and after the imiting reagent is consumed during the The excess reagent was Sos reaction, the mass of excess reagent remaining will be 7.25 g 12.3g 0.1531 SO2 0.03 g 1.45 g 840 PM 11/7/2019 SP S

Answer 1

Balanced equation

2SO2 + O2 ----> 2SO3

Number of moles = mass in gram / Molar mass

Molar mass of SO2 = 64.07 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of SO3 = 80.07 g/mol

Number of moles of SO2 = 5.80 g / 64.07 g/mol = 0.0905 mol

Number of moles of O2 = 2.45 g / 32 g/mol = 0.0766 mol

From reaction, 2.0 mol of SO2 reacts with 1.0 mol of O2 produces 2.0 mol of SO3 then 0.0905 mol of SO2 will react with 1.0 mol * 0.0905 mol / 2.0 mol = 0.04525 mol of O2 will produce 0.0905 mol of SO3.

Since SO2 is reacting completely during the reaction so SO2 is limiting reagent.

Number of moles of SO3 formed = 0.0905 mol

Mass of SO3 produced = 0.0905 mol * 80.07 g/mol = 7.25 g

Since O2 is not reacting completely during the reaction so O2 was excess reagent.

Number of excess reagent O2 remaining = (0.0766 - 0.04525) mol = 0.03135 mol

Mass of excess reagent O2 remaining = 0.03135 mol * 32 g/mol = 1.00 g