A 1. 0.10g of CH 4 with 0.50g of O 2 2. Heat of combustion of CH 4 590KJ/mol 3.
ID: 1005452 • Letter: A
Question
A 1. 0.10g of CH4 with 0.50g of O2
2. Heat of combustion of CH4 590KJ/mol
3. Fe (III) 416 nm
4. P
5. HCN
Q1. For the given hydrocarbon and molecular oxygen masses, provide the relevant balanced combustion equation. With this and the given masses, determine the Limiting reagent, the Excess Reagent, the theoretical yield of CO2, the actual and % yields. (40pts)
Q2. For the given heat of combustion, determine the heat produced for a 0.10g of the given compound. (10pts)
Q3. For the given Atomic Absorption emission wavelength of Fe (III), using Planck’s equation, determine the energy associated with this emission. Given the wavelength, what color is associated with this emission? Given the energy, do you think this is a single step (example n=3 to n=2) or a multi step (example n=4 to n=2)? (10pts)
Q4. For the given element, provide the correct orbital diagram, electron configuration and give the number of valence electrons (20pts)
Q5. For the given molecular formula, give the FULL Lewis Structure (indicating valence calculation), the VSEPR geometry including a 3-D drawing, and indicate polarity of bonds and the overall molecule. (30pts)
Explanation / Answer
1 ) CH4 + O2 ---> CO2 + 2H2O
no of mol of CH4 = 0.1/16 = 0.00625 mol
NO of mol of O2 = 0.5/32 = 0.0156 mol
limiting reagent = CH4
EXCESS REAGENT = o2
Theoretical yield of CO2 = 0.00625*44 = 0.275 grams
actual yield = experimental data not given.
%yield = actual yield / Theoretical yield*100
2) DHrxn = 590 kj/mol
heat of combustion = 590*0.00625 = 3.6875 kj
3) E = hv
= 6.625*10^-34*(3*10^8/(416*10^-9))
= 4.77*10^-19 joule
from n=4 to n=2
DE = 2.18*10^-18(1/2^2-(1/4^2))
= 4.085*10^-19 joule.
from n=3 to n=2
DE = 2.18*10^-18(1/2^2-(1/3^2))
= 3.027*10^-19 joule.
based on the above data it is multi step
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