Question

Complete the following reaction: $C_{3}H_{8}(g)+3H_{2}O(g)\xrightarrow[\text{Catalyst}]{\Delta}$

Answer 1

The reaction of propane and water is not a natural process.

But under very high temperature (as high as 1200-1300 degree centigrade) and with catalyst this reaction can take place to form methyl ketene, also known as propargyl alcohol and hydrogen.

In this reaction water is actually in the form of steam due such high temperature, the reaction is as follows;

$C_3H_8(g)+3H_2O\xrightarrow [Heat]{catalys} C_3H_4O(g)+3H_2(g)$

Answer 2

Answer:

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