Question #6be2c

1 Answer
Jul 25, 2017

Answer:

This is an ill-stated problem.

Explanation:

This is an ill-stated problem. It is not explicit whether the water product is condensed out after the explosion, or is removed with the carbon dioxide in the NaOH wash. It also does not specify the form of the hydrocarbon. At a minimum it should state whether it is expected to be an alkane, an alkene or an alkyne, assuming that it is NOT any oxygenated organic compound! It also does not state whether any diluent gas is present (e.g. N2) or if all of the non-HC original gas is oxygen.

With excess oxygen we know that the oxidation is complete, so the entire hydrocarbon turned into carbon dioxide and water. In this instance we can assume that the gases exhibit close to ideal gas behavior, so the volumes are equivalent to molar quantities. From the reactions given we know that the relative molar amounts of carbon dioxide and water is 30, the original hydrocarbon is 10, and the reduction of oxygen – what was consumed – is 16.
#H_(n+2)C_n + (((n+2)/2)+n)O_2 → nCO_2 + ((n+2)/2)H_2O + xO_2#

Minimum original volume (moles) must be the sum of the decreases – 55. 10 for the HC leaves 45 for the oxygen. Assuming final products are all in the gas phase, the 10 volumes of HC must produce 10 volumes of #CO_2# and thus 6 of #H_2O# while consuming (depleting) 16 #O_2#. That means that only the HC volume was lost, or 10, not 25.

Even total water condensation at that point would only reduce the volume by another 6 volumes. But, that would mean the #CO_2# produced was 30 volumes. It doesn’t balance.