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Tutor profile: Madeline C.

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Madeline C.
Organic Chemistry PhD Student
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Questions

Subject: Organic Chemistry

TutorMe
Question:

Consider the following reaction: CH3CH2Cl + NaOH -> ? Determine the reaction mechanism and the product.

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Madeline C.
Answer:

To determine the mechanism, we need to consider several pieces of the reaction: the leaving group, the electrophile, the nucleophile, and any possible steric hindrance. Cl is our leaving group, and because of its size it is a good leaving group. The methylene group (-CH2-) is our electrophile due to the dipole created by the C-Cl bond. OH- is our nucleophile and its negative charge makes it a string nucleophile. Both molecules are relatively small and our leaving group is primary, so steric hindrance will not be an issue in this problem. With a primary leaving group and a strong nucleophile, we can conclude that this reaction will proceed through an SN2 mechanism. The OH will attack the methylene and the Cl will be pushed off. This leaves us with the following reaction: CH3CH2Cl + NaOH -> CH3CH2OH + NaCl

Subject: Chemistry

TutorMe
Question:

Consider the following reaction: CO(g) + 2 H2(g) -> CH3OH(l) Assume you have 300g CO and 50g H2. How many grams of CH3OH can you make?

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Madeline C.
Answer:

First, we need to determine what the limiting reagent in the problem is. The limiting reagent is the one that will run out the fastest once the reaction starts. This limits how much product can be made. To determine this, we need to know how many moles of each reagent we have. We can figure this out using stoichiometry and the molar masses of each of our compounds: CO is 28.01 g/mol, H2 is 2.016 g/mol 300g CO / 28.01 g/mol = 10.71 mol CO 50g H2 / 2.016 g/mol = 24.80 mol H2 We also need to remember that our equation uses 2 mol H2 for every 1 mol CO, so our final numbers are 10.71 mol CO and 12.40 mol (2 H2). We have fewer moles of CO, so it will run out faster during the reaction. This makes CO the limiting reagent and therefore this is the number we need to look at when calculating the total amount of product. With 10.71 mol of our limiting reagent, we can make a maximum of 10.71 moles of product. To convert this to grams, we’ll use stoichiometry again: CH3OH is 32.04 g/mol 10.71 mol CH3OH x 32.04 g/mol = 343.15 g CH3OH

Subject: Basic Chemistry

TutorMe
Question:

Balance the following equation: Al + O2 -> Al2O3

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Madeline C.
Answer:

The first thing we need to know about balancing equations is that there needs to be the same number of each atom on both sides of the arrow. However, we also want the lowest total number of atoms possible. Here we have 1 Al and 2 O on the left and 2 Al and 3 O on the right. Let’s start with the O: we need a number that is divisible by both 2 and 3 to balance these. 6 is the lowest number that we can use so we need 6 O on each side. To make that happen, we put coefficients in front of the O containing compounds to get this: Al + 3 O2 -> 2 Al2O3 Now our Os are balanced, but we still need to balance the Al. Right now we have 1 Al on the left and 4 Al on the right. Since Al is not part of a compound on the left, we can simply stick a coefficient in there to make the number match: 4 Al + 3 O2 -> 2 Al2O3 Now we have 4 Al and 6 O on the left and 4 Al and 6 O on the right. Our equation is balanced!

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