Q12 V2A compound underwent two tests.In the first test, 1 mole of the compound was completely combusted, yielding 3 moles of CO2.In the second test, the compound did not show any rapid reaction with bromine. The formula of the compound is likely to be:

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BlackTom AI · Accepted Answer

To begin, I restate the information given: in the first combustion test, 1 mole of the compound yields 3 moles of CO2 when fully burned. In the second test, the compound does not show any rapid reaction with bromine.

Option a: C3H6. If one mole of this compound were combusted completely, the number of CO2 moles produced equals the number of carbon atoms in the molecule, since each carbon ends up as CO2. C3H6 has 3 carbons, so we would expect 3 moles of CO2, which matches the observed 3 CO2. However, C3H6 is unsaturated (has at least one C=C or C≡C bond). When bromine water is added to an unsaturated hydrocarbon, the bromine adds across the double bond, causing a rapid decolorization. The given second test states there was no rapid reaction with bromine, which argues against an unsaturated hydrocarbon.

Option b: C2H4. This molecule has 2 carbons, so complete combustion would produce about 2 moles of CO2, not 3. Therefore, this option cannot match the first test.

Option c: C4H10. This alkane has 4 carbons, so complete combustion would yield 4 moles of CO2 per mole of fuel, again not matching the observed 3 CO2. So this option is inconsistent with the first test.

Option d: C3H8. This is a saturated hydrocarbon (propane) with 3 carbons. Upon complete combustion, it would indeed yield 3 moles of CO2 per mole of fuel, aligning with the first test.

Now, considering the second test: the compound did not show any rapid reaction with bromine. Saturated hydrocarbons like alkanes (e.g., C3H8) do not react readily with bromine under normal conditions, whereas unsaturated hydrocarbons (like C3H6) react rapidly due to the presence of carbon–carbon multiple bonds. Therefore, the lack of a rapid bromine reaction further supports a saturated formula.

In summary, the requirement from the first test intersects with the requirement from the second test: a 3-carbon, saturated hydrocarbon matches both observations. The combination of these two pieces of evidence narrows the possibilities to a propane-type molecule that is fully saturated, which is consistent with C3H8.

Q12 V2A compound underwent two tests.In the first test, 1 mole of the compound was completely combusted, yielding 3 moles of CO2.In the second test, the compound did not show any rapid reaction with bromine. The formula of the compound is likely to be:单项选择题

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