A reaction was set up to convert an alcohol to a carbonyl. Samples of the reaction mixture were checked by infrared spectrometry every couple of minutes. Use the infrared spectra provided to determine if the reaction mixture contains: pure alcohol, a mixture of alcohol and carbonyl, or pure carbonyl. The reaction mixture contains pure carbonyl. pure alcohol. mixture of alcohol and carbonyl. The reaction mixture contains mixture of alcohol and carbonyl. pure carbonyl. pure alcohol.

First spectrum = contains only alcohol Compound.

Second spectrum= Contains both alcohol and carbonyl compounds.

Explanation:

The concept known as "spectroscopy'' is an important aspect of Chemistry. It is used in the identification of chemical compounds. Some of the techniques used in spectroscopy is through the use of NMR, IR spectroscopy and UV-Vis spectroscopy.

This question is about IR spectroscopy. Here each functional group has its own unique frequency of absorption. For instance the frequency of absorption of Alkane, C- H( stretch) is 2850 - 3000, alkyne, C-H( stretch) is 3300, amine, N-H(bending) is 1600 and so on.

For the characteristic frequency of absorption found in OH(stretch) is 3300 - 3700. This can be seen in the spectrum given. Thus, OH bond is present.

For carbonyl group, C= O the characteristic absorption frequency is 1670-1820. This frequency can not be found in the first spectrum. But checking the second spectrum a frequency of 1700 can be seen.

Thus, the C=O bond is absent on the first spectrum but present in the second spectrum.

In the second spectrum too, the characteristic frequency of absorption of 3300 is found. Thus, in the second spectrum the compound have both OH and C=O.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

you need to have a picture of the atom that way we can answer it