Please find the CO2 amount and show the calculation: Air composition (apprx) = 80% N2 and 20% O2 (mol basis only) Average molecular weight of Air = (0.8 x 28 + 0.2 x 32) = 29 g/mol (apprx) Math Sample: 2kg pure charcoal (12C) is to be burnt completely with air. Find the air, CO2 and N2 amount in kg’s. Solution: C + O2 = CO2 (you must use a balanced equation) Therefore from the mole ratio of the reaction we write, C : O2 : CO2 = 1:1:1 and from air composition we got O2 : N2 : Air = 1:4:5 Given, 2 kg C = 2000 g/ (12 g/mol) = 166.67 mole C Therefore, similar mole of O2 required. So equivalent Air supposed to be 5 times than the mole amount of O2 and released N2 will be 4 times than the required O2. Therefore, Air amount = 5 x 166.67 moles = 833.34 moles = (833.33x29/1000) kg = 24.17 kg air Similarly N2 released amount will be = (166.67 x 4 x 28/1000) kg = 18.67 kg Find CO2 amount by yourself! (Isn’t it 7.34 kg?)
Please find the CO2 amount and show the calculation: Air composition (apprx) = 80% N2 and 20% O2 (mol basis only) Average molecular weight of Air = (0.8 x 28 + 0.2 x 32) = 29 g/mol (apprx) Math Sample: 2kg pure charcoal (12C) is to be burnt completely with air. Find the air, CO2 and N2 amount in kg’s. Solution: C + O2 = CO2 (you must use a balanced equation) Therefore from the mole ratio of the reaction we write, C : O2 : CO2 = 1:1:1 and from air composition we got O2 : N2 : Air = 1:4:5 Given, 2 kg C = 2000 g/ (12 g/mol) = 166.67 mole C Therefore, similar mole of O2 required. So equivalent Air supposed to be 5 times than the mole amount of O2 and released N2 will be 4 times than the required O2. Therefore, Air amount = 5 x 166.67 moles = 833.34 moles = (833.33x29/1000) kg = 24.17 kg air Similarly N2 released amount will be = (166.67 x 4 x 28/1000) kg = 18.67 kg Find CO2 amount by yourself! (Isn’t it 7.34 kg?)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please find the CO2 amount and show the calculation:
Air composition (apprx) = 80% N2 and 20% O2 (mol basis only)
Average molecular weight of Air = (0.8 x 28 + 0.2 x 32) = 29 g/mol (apprx)
Math Sample:
2kg pure charcoal (12C) is to be burnt completely with air. Find the air, CO2 and
N2 amount in kg’s.
Solution:
C + O2 = CO2 (you must use a balanced equation)
Therefore from the mole ratio of the reaction we write,
C : O2 : CO2 = 1:1:1
and from air composition we got
O2 : N2 : Air = 1:4:5
Given, 2 kg C = 2000 g/ (12 g/mol) = 166.67 mole C
Therefore, similar mole of O2 required. So equivalent Air supposed to be 5
times than the mole amount of O2 and released N2 will be 4 times than the
required O2.
Therefore, Air amount = 5 x 166.67 moles
= 833.34 moles
= (833.33x29/1000) kg
= 24.17 kg air
Similarly N2 released amount will be = (166.67 x 4 x 28/1000) kg = 18.67 kg
Find CO2 amount by yourself! (Isn’t it 7.34 kg?)
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