Answer:
a) ethanol is dehydrated by concentrated sulphuric acid to yield ethene
b) Concentrated phosphoric(V) acid, H3PO4, can be used instead of sulphuric acid
Explanation:
Ethene is prepared in the lab by heating ethanol with concentrated sulphuric acid at 170°C. The mixture contains 1:2 volume ratio of the ethanol and concentrated acid. Ethylhydrogentetraoxosulphate VI is first formed and then decomposes to yield ethene and sulphuric acid.
The overall reaction is akin to the dehydration of ethanol. As is easily seen above; this is a two stage reaction. The first stage is the formation of the ester, the second stage involves the dehydration of the ester to yield ethene. The ethene gas is passed through concentrated sodium hydroxide to remove any gaseous impurities present and then it is collected over water.
Concentrated phosphoric(V) acid, H3PO4, can be used instead of concentrated sulphuric acid as a dehydrating agent in the laboratory preparation of ethene.
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Set up a proportion os V1/n1 = V2/n2. V is the volume, n is the amount in MOLES, not grams. Convert the CO_2 to moles, then solve and find that the amount of N_2 should be the same amount of moles. Then use the molar mass of N_2 (28.02 grams/mole) to convert that amount of moles into grams. That's your answer.
