The carbon dioxide exhaled by astronauts can be removed from a spacecraft by reacting it with lithium hydroxide (LiOH). The reac
1 answer:
You might be interested in
Answer:
See answer for details
Explanation:
In this case, we have the following substract:
Ph-CH2 - O - CH2CH3
That's the benzyl ethyl ether.
When this compound reacts with a concentrated Solution of HI, as we have an acid medium, we will have an SN1 reaction, where the Hydrogen will attach to the oxygen and the iodine will go to the most stable carbon, in this case, the carbon next to the ring is the most stable (because of the double bond of the ring, charges can be stabilized better this way), so product A will be the benzyl with the iodine and product B, will be the alcohol:
A: Ph - CH2 - I
B: CH3CH2OH
When B reacts again with HI, it's promoting another SN1 reaction, where the OH substract the H from the HI, the OH2+ will go out the molecule, leaving a secondary carbocation (CH2+) and then, the Iodine (I-) can go there via SN1 and the final product would be:
C: CH3CH2I
See picture for mechanism:
Answer:
127 grams of carbon dioxide
Explanation:
We need to determine the chemical equation first. Butane has a chemical formula of , oxygen is
, carbon dioxide is
, and water is
. The reactants are butane and oxygen and the products are carbon dioxide and water. So we write:
⇒
But remember! We need to balance this. Currently, there are 4 carbon atoms (C), 10 hydrogen atoms (H), and 2 oxygen atoms (O) on the left, while there are 1 carbon atom (C), 2 hydrogen atoms (H), and 3 oxygen atoms (O) on the right. Let's place a coefficient of 4 in front of the carbon dioxide and a coefficient of 5 on the water, so that we have equal numbers of carbon and hydrogen atoms on each side:
⇒
However, we need to ensure that there are equal numbers of O atoms, as well. On the left, we have 2 and on the right we have 13, so let's put a coefficient of 6.5 on the oxygen:
⇒
Finally, multiply everything by 2 to get whole number coefficients:
⇒
Ah, now we can actually get to the problem!
We need to determine the limiting reactant, so let's convert the 42 g of butane and 150 g of oxygen into moles of any product, say, carbon dioxide. To convert to moles, we need to find the molar mass of each compound.
The molar mass of butane is 4 * 12.01 + 10 * 1.01 = 58.14 g/mol, while the molar mass of oxygen is 2 * 16 = 32 g/mol. We can now set up the equations:
Clearly, we see that 2.8846 < 2.8896, which means that oxygen is the limiting reactant. In other words, the most products can be made when the oxygen is all used up.
Now let's finally convert moles of carbon dioxide into grams by multiplying by its molar mass, which is 12.01 + 2 * 16 = 44.01 g/mol:
Notice that we have 3 significant figures because we had 3 significant figures at the start with 150. grams of oxygen.
<em>~ an aesthetics lover</em>