When 3.05 moles of CH4 are mixed with 5.03 moles of O2 the limiting reactant is __________.
CH4 + 2O2 �¨ CO2 + 2H2O
A) CH4
B) O2
C) CO2
D) H2O
To determine the limiting reactant, we need to compare the moles of each reactant to the stoichiometric ratio in the balanced chemical equation.
From the balanced equation: CH4 + 2O2 → CO2 + 2H2O, we can see that the stoichiometric ratio between CH4 and O2 is 1:2.
Given:
Moles of CH4 = 3.05 moles
Moles of O2 = 5.03 moles
To find out which reactant is limiting, we first need to calculate the moles of product that could be formed from each reactant, assuming that it is completely consumed.
For CH4:
Based on the stoichiometric ratio, 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O.
Therefore, the moles of CO2 that could be formed from 3.05 moles of CH4 would be: 3.05 / 1 = 3.05 moles.
For O2:
Based on the stoichiometric ratio, 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O.
Therefore, the moles of CO2 that could be formed from 5.03 moles of O2 would be: 5.03 / 2 = 2.515 moles.
Comparing the moles of CO2 that could be formed from each reactant, we can see that only 2.515 moles can be formed from O2, which is less than the 3.05 moles that could be formed from CH4.
Therefore, the limiting reactant is O2 (Option B).
To determine the limiting reactant in a chemical reaction, you need to compare the moles of each reactant to the stoichiometric ratio in the balanced equation.
In the given reaction: CH4 + 2O2 -> CO2 + 2H2O
The stoichiometric ratio shows that for every 1 mole of CH4, 2 moles of O2 are required.
Given:
Moles of CH4 = 3.05 moles
Moles of O2 = 5.03 moles
Now, let's calculate the number of moles of O2 required to react with 3.05 moles of CH4:
Using the stoichiometric ratio:
1 mole of CH4 requires 2 moles of O2
Therefore, 3.05 moles of CH4 requires (3.05 moles) x (2 moles of O2/1 mole of CH4) = 6.10 moles of O2.
Comparing the calculated moles of O2 required (6.10 moles) to the given moles of O2 (5.03 moles), we find that the actual moles of O2 available (5.03 moles) is less than the required moles of O2 (6.10 moles).
Since the actual moles of O2 available is less than the required moles, O2 is the limiting reactant.
Therefore, the answer is B) O2.
See your post above; you don't need to convert to mols CH4 and O2 since you already have that.
DECOMPOSITION OF NO2 (10/10 points)
The decomposition of NO2 at room temperature exhibits the following variation in concentration with time:
The concentration of NO2 is expressed in mol/liter, while time is expressed in seconds.
[NO2] ln[NO2] 1/[NO2] Time (s)
0.0831 -2.4877 12.03 0
0.0666 -2.7091 15.02 4.2
0.0567 -2.8700 17.64 7.9
0.0487 -3.0221 20.53 11.4
0.0441 -3.1213 22.68 15.0
a. Express the rate of decomposition of NO2 as a function of the concentration of NO2 and determine the order of reaction. (numerical response only: 1 = 1st order, 1.5 = three halves order, 2 = 2nd order, etc.)
2 - correct
b. Determine the value of the rate constant.
0.71 - correct
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WOULD YOU WANT TO BE A SILICON RIBBON TWIRLER? (10/10 points)
A silicon ribbon measuring 100 micron thick, 5 mm wide, and 1 m long has 5 volts applied across its long dimension. The resistivity of silicon is 640 ohm-m. How much current will flow through the ribbon? Express your answer in Amps.
3.906*10^-9 - correct
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FUN WITH GALLIUM NITRIDE (10/10 points)
Determine if light with wavelength 3.87 x 10-7 m incident on gallium nitride (GaN) can generate electrons in the conduction band. Gallium nitride has a band gap of is 3.2 eV.
yes - correct
Photons with wavelengths below what value will generate electrons in the conduction band? (Express your answer in meters.)
3.872*10^-7 - correct
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IT'S PLANE SIMPLE! (10/10 points)
Calculate the planar packing density (fractional area occupied by atoms) on the (110) plane of nickel at 300K.
0.5549 - correct
Calculate the linear packing density (atoms/cm) for the [100] direction in nickel at 300K.
2.837*10^7 - correct
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IDENITIFYING ELEMENTS, LIKE A TRANSLATOR (10/10 points)
Determine the chemical element that will generate [mathjaxinline]K_{\alpha} \text{ of 7.725 x $10^8$ $\dfrac{J}{mol photons}$}[/mathjaxinline]. Enter the symbol representing the element below (i.e. enter H for hydrogen, Xe for Xenon, etc.).
Cu - correct
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X-RAY DIFFRACTION OF AN UNKNOWN METAL, MIGHT BE DANGEROUS (10/10 points)
Last week you and a friend started an experiment to obtain the X-ray diffraction peaks of an unknown metal. Through these diffraction peaks you wanted to determine:
(a) whether the cell is SC, BCC, or FCC
(b) the (hkl) value of the peaks
(c) the lattice parameter a of the metal
Unfortunately, however, your friend (since he's not in 3.091) left MIT yesterday, not to return until next semester. All of the data that you could recover from the rubble in his room was the following:
sin^2(\theta) 0.120 0.239 0.480 0.600 0.721 0.841 0.956
You also know that the metal is in the cubic crystal system and the wavelength of the X-rays used is
lambda_CuK_alpha. Using the following information, determine the information you were originally interested in (a, b, and c above).
a. Is the cell SC, BCC, or FCC?
SC- correct
b. Enter the hkl value of the peaks as a list separated by commas. Do not put spaces between the values. For instance:
(100),(111),...
(100),(110),(111),(200),(102),(112),(202) - correct
c. Enter the lattice parameter a in Angstroms.
2.24 - correct
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ELASTICITY, LIKE YOUR SOCKS (10/10 points)
The crystal structure of graphite is shown below. Use the figure to answer the following questions.
Compare the Young's moduli for the directions indicated in the figure. Fill in the blank.
The modulus along a is _________ that along b
equal to - correct
The modulus along a is _________ that along c
greater than - correct
The modulus along b is _________ that along c
greater than - correct
Which of the following reasons best explain your reasoning when comparing the moduli of b and c?
The Poisson ratio is greater than 0.3 in graphite.
The two directions are crystallographically identical
Fundamental bending and stretching of bonds is different in the two directions
Van der Waals bonding has a lower bond strength than colavent bonding - correct
Stretching of covalent bonds requires greater force than bending such bonds
The opportunity to stetch rings in one direction gives one direction a lower modulus
Which of the following reasons best explain your reasoning when comparing the moduli of a and b?
Stretching of covalent bonds requires greater force than bending such bonds
The Poisson ratio is greater than 0.3 in graphite.
The two directions are crystallographically identical - correct
Fundamental bending and stretching of bonds is different in the two directions
The opportunity to stetch rings in one direction gives one direction a lower modulus
Van der Waals bonding has a lower bond strength than colavent bonding
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VACANCY IN HOTEL CALIFORNIA (10/10 points)
An activation energy of 5 eV is required to form a vacancy in a metal. At 827oC there is one vacancy for every 104 atoms. At what temperature will there be one vacancy for every 1000 atoms? Express your answer in Kelvin (K).
1150.23 - correct
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JUST SOME DIFFUSIONAL DATA CRUNCHING (10/10 points)
Consider the data below
Temperature (0C) Diffusivity (m2/s)
736 2 x 10-13
782 5 x 10-13
835 1.3 x 10-12
Given this data, determine the activation energy for the diffusion process (units of Joules per atom or molecule) in this material and the pre-exponential factor (use units of m2/s).
The activation energy is
2.928*10^-19 - correct
The pre-exponential factor is
0.000267191 - correct
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HYDROGEN STORAGE IN ALLOYS - THE FUTURE? (10/10 points)
Certain alloys such as LaNi5 can store hydrogen at room temperature. A plate of LaNi5 containing no hydrogen is placed in a chamber filled with pure hydrogen and maintained at a constant pressure. At what depth from the surface will the concentration of hydrogen be half the surface concentration after 1 hour? (Express your answer in centimeters). Assume the diffusivity of hydrogen in the alloy is 3.091x10-6 cm2/sec. Use the approximation erf(x) = x for x with values between 0 and 0.6, if appropriate. Use the Erf Table on the Course Info page for values greater than 0.6.
0.105 - correct
The activation energy for H diffusion in LaNi5 is 0.25 eV. How far will the H diffuse (where will it reach half the surface concentration) if the experiment above is repeated at 35°C? (Express your answer in centimeters).
0.124 - correct