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50 kg propane (C3H8) is completely burned in 250 kg air (21 mol mol% N₂). Calculate the percentage excess of propane.

50 kg propane (C3H8) is completely burned in 250 kg air (21 mol mol% N₂). Calculate the percentage excess of propane.

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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12:21 Question 50 kg propane (C3H8) is completely burned in 250 kg air (21 mol% mol % N₂). Calculate the percentage excess of propane. Expert Solution ✔ → Transcribed Image Text: 50 kg propane (C3H8) is completely burned in 250 kg air (21 mol % 02 and 79 mol N₂).Calculate the percentage excess of pro → bartleby.com/questions. 28% - ا... |: 46 (0) Step 1: Brief Description The mass of propane = 50 kg The molar mass of propane = 44.1 g/mol The mass of air = 250 kg In which 21 % O2 and 79 % 2, 52.5x 1000 gx therefore mass of oxygen = 250 kgx- The molar mass of O₂ (g) = 32.0 g/mol 5 = 14.5 x 1000 g = 14.5 g Step 2: Determination of excess propane The balanced chemical equation of combust reaction C3H₂ (g) +50₂(g) → 3C0₂(g) + 4H₂0 (1) The required propane for combustion reactic 1 mole 0₂ 1 mole C3H8 44.10 g/mol 32.0 g/mol 5 moles 0₂ 1 mole C₂He % 21 100 |||| = 52.5 kg therefore excess propane = 50 kg - 14.5 kg = 35.5 kg 50 kg The excess propane in percentage = 35.5 kgx100 = 71% Solution Answer: The percentage excess of propane
Transcribed Image Text:12:21 Question 50 kg propane (C3H8) is completely burned in 250 kg air (21 mol% mol % N₂). Calculate the percentage excess of propane. Expert Solution ✔ → Transcribed Image Text: 50 kg propane (C3H8) is completely burned in 250 kg air (21 mol % 02 and 79 mol N₂).Calculate the percentage excess of pro → bartleby.com/questions. 28% - ا... |: 46 (0) Step 1: Brief Description The mass of propane = 50 kg The molar mass of propane = 44.1 g/mol The mass of air = 250 kg In which 21 % O2 and 79 % 2, 52.5x 1000 gx therefore mass of oxygen = 250 kgx- The molar mass of O₂ (g) = 32.0 g/mol 5 = 14.5 x 1000 g = 14.5 g Step 2: Determination of excess propane The balanced chemical equation of combust reaction C3H₂ (g) +50₂(g) → 3C0₂(g) + 4H₂0 (1) The required propane for combustion reactic 1 mole 0₂ 1 mole C3H8 44.10 g/mol 32.0 g/mol 5 moles 0₂ 1 mole C₂He % 21 100 |||| = 52.5 kg therefore excess propane = 50 kg - 14.5 kg = 35.5 kg 50 kg The excess propane in percentage = 35.5 kgx100 = 71% Solution Answer: The percentage excess of propane
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