which conversion factor do you use first to calculate the number of grams of CO2 produced by the reaction of 50.6 gram of CH4 with O2? The equation for the complete combustion of methane is: CH4+2O2-->CO2+2H2O.
Unit 6 Lesson 2: Chemical Calculations
1. Calculate the number of moles of Al2O3 that are produced when 0.60 mol of Fe is produced in the following reaction.
2Al(s) + 3FeO(s) --> 3Fe(s) + Al2O3(s)
A, 0.20 mol
2. At STP, how many liters of oxygen are required to react completely with 3.6 liters of hydrogen to form water?
2h2(g) + O2(g) --> 2H2O(g)
A, 1.8 L
3. Which conversion factor do you use first to calculate the number of grams of CO2 produced by the reaction of 50.6g of CH4 with O2? The equation for the complete combustion of methane is: CH4(g) + 2O2(g) --> CO2(g) + 2H2O (l)
A, 1 mol CH4/16.0g CH4
4. Which of the following statements is true about the following reaction?
3NaHCO3(aq) + C6H8O7(aq) --> 3CO2(g) + 3H2O(s) + Na3C6H507(aq)
B, 1 mole of water is produced for every mole of carbon dioxide produced.
First convert 50.6 g CH4 to mols.
mols CH4 = 50.6 g CH4 x (1 mol CH4/16 g CH4) = ??
it si 1mol CH4/16.0 g CH4
To calculate the number of grams of CO2 produced by the combustion reaction of 50.6 grams of CH4, you would use the conversion factor that relates the amount of CH4 to the amount of CO2 produced.
Based on the balanced equation for the reaction, we can see that 1 mole of CH4 reacts to produce 1 mole of CO2. This means that the conversion factor is 1 mole of CH4/1 mole of CO2.
To use this conversion factor, follow these steps:
Step 1: Determine the moles of CH4. To do this, divide the given mass of CH4 by its molar mass.
The molar mass of CH4 is:
(1 x atomic mass of C) + (4 x atomic mass of H) = (1 x 12.01 g/mol) + (4 x 1.01 g/mol) = 16.05 g/mol
So, the moles of CH4 = 50.6 g CH4 / 16.05 g/mol = 3.16 moles of CH4
Step 2: Apply the conversion factor. Since the conversion factor is 1 mole of CH4/1 mole of CO2, you can multiply the moles of CH4 by the conversion factor to obtain the moles of CO2.
Moles of CO2 = 3.16 moles of CH4 x (1 mole of CO2/1 mole of CH4) = 3.16 moles of CO2
Step 3: Convert moles of CO2 to grams. To do this, multiply the moles of CO2 by its molar mass.
The molar mass of CO2 is:
(1 x atomic mass of C) + (2 x atomic mass of O) = (1 x 12.01 g/mol) + (2 x 16.00 g/mol) = 44.01 g/mol
So, the grams of CO2 = 3.16 moles of CO2 x 44.01 g/mol = 139.28 g CO2
Therefore, the number of grams of CO2 produced by the reaction of 50.6 grams of CH4 is 139.28 grams.
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