ME300Lecture21

ME300Lecture21 - Lecture 21 : Stoichiometry of Reactions...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
21.1 Lecture 21 : Stoichiometry of Reactions Objective: » Systems involving chemical reactions » Chemical composition changing during a process Combustion Process: » The rapid oxidation of combustible elements of fuel for energy release: Fuel + Oxidizer Exhaust Gases (Reactants) (Products) » Usually requires “ignition”
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
21.2 Continue Combustion Process Fuels: » Major combustible chemical elements in common fuels: Carbon, Hydrogen, and Sulfur » Most common fuel : Hydrocarbons, C x H y » e.g., Substance chemical formula methane CH 4 propane C 3 H 8 octane C 8 H 18
Background image of page 2
21.3 Continue Combustion Process Combustion Air (Oxidizer) » In most combustion processes, air provides the needed oxygen, although pure oxygen (O 2 ) would be best » Modeling of combustion air: – Air only consists of 21% oxygen and 79% nitrogen on a molar basis (by volume) molar ratio of N 2 to O 2 = 3.76 1 mol O 2 + 3.76 mols N 2 = 4.76 mols air – Air is considered on a dry basis (no moisture) – Nitrogen is inert (does not participate in the reaction) Conservation of Mass » Mass of reactant elements = Mass of product elements
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
21.4 Example 21.1 Given: H 2 + 0.5 O 2 H 2 O this really means, 1 mol H 2 + 0.5 mols O 2 1 mol H 2 O Note: Moles are not conserved, but mass is! 1 mol H 2 = 2 mols H = 2 mols x1 g/mol = 2 g 0.5 mol O 2 = 1 mol O = 1 mol x 16 g/mol = 16 g 1 mol H 2 O = 2 mol H + 1 mol O = 18 g
Background image of page 4
21.5 Example 21.2 Given: H 2 + 0.8 (O 2 + 3.76 N 2 ) a H 2 O + b O 2 + c N 2 Find: Amounts of product gases Solution: » Proceed by balancing moles of each element Hydrogen (H) balance: 2 = 2a a = 1
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 04/23/2011.

Page1 / 18

ME300Lecture21 - Lecture 21 : Stoichiometry of Reactions...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online