Please do #25 and #26 with explanations. Dismiss anything I\'ve already written
ID: 531872 • Letter: P
Question
Please do #25 and #26 with explanations. Dismiss anything I've already written on the paper since I have no idea what I'm doing. I also don't understand how/if oxygen being a diatomic molecule influences this problem. I use the railroad method to figure these out (what I did on the right there for #25 so I would appreciate if you could also do it that way if you are able to). I am a visual learner so that way is best for me. Thank you!! Questions 25 & 26 involve the following reaction: produce? CH 102 C. 22g B. 11g D. 811g mol (01 443 Ilog CH4 Give the reasoning for your answer show your work). Imol 0 mol 0 26. If you obtained 5 gco, what would be the percent yieldn 7. yittd actual A. 11.4% C. 45.5% B. 22.7% D. 62.5% 27. Which diagram best shows hydrogen at 20K (in the liquid phase)? 28. The boiling point of hydrogen is 20K. Describe the particles present in a container of hydrogen at 25K. A. Gaseous hydrogen atoms B. Gaseous hydrogen molecules C. Hydrogen atoms in the liquid state D. Hydrogen molecules in the liquid state
Explanation / Answer
25. CH4 + 2O2 ------> CO2 + 2H2O
Molar mass of methane = 16g/mol
Molar mass of Oxygen = 32 g/mol
Mass of methane present = 16 g So moles of methane present = mass of methane/molar mass of methane = 16/16 = 1 mol.
Mass of Oxygen present = 16 g. So moles of oxygen present = mass of oxygen/molar mass of oxygen = 16/32 = 0.5 moles
From the reaction stoichiometry, it is seen that 1 mole of methane requires 2 moles of Oxygen to produce 1 mole of CO2 and 2 moles of H2O.
But we have only 0.5 moles of Oxygen present, so not all methane will get reacted but only a part of it corresponding to 0.5 moles of oxygen.
2 moles of oxygen require 1 moles of methane to react. So, moles of methane required to react with 0.5 moles of Oxygen = 0.5 x (1/2) = 0.25
Therefore, out of 1 mole of Methane present, only 0.25 moles will react.
1 mole of methane produces 1 mole of Carbon dioxide. So, 0.25 moles of methane will produce Carbon dioxide = 0.25 x (1/1) = 0.25 moles
Molar mass of Carbon dioxide = 44 g/mol
So, mass of carbon dioxide present in 0.25 moles = 0.25 x 44 = 11 grams. Option B is correct.
26. Actual amount of Carbon dioxide produced = 5 grams.
From question 25 above, it is seen that with 16 grams of methane and 16 grams of oxygen should theoretically produce 11 grams of Carbon dioxide. Percent yield = (actual amount of CO2 produced/theoretical amount of CO2 that should be produced) x 100 = (5/11) x 100 = 45.45 % (approximately = 45.5 %) Option C is correct