# How much volume of oxygen will be required for complete combustion of 40 ml of acetylene and how much volume of CO2 and H2O will be formed?

How much volume of oxygen will be required for complete combustion of 40 ml of acetylene and how much volume of $\ce{CO2}$ and $\ce{H2O}$ will be formed?

After balancing the left-hand side produces 5 volume of oxygen and 2 volume of acetylene. However, the right-hand side produces 4 volume of carbondioxide and 2 volume of water. Why doesn't the volume on LHS and RHS match? Or how is it matched?

Is it correct that 100 ml of oxygen and 40 ml of acetylene produce 80 ml of $\ce{CO2}$ and 40 ml of $\ce{H2O}$? If so, what happened to 20 ml on the right hand side?

• I think this shouldn't be on hold. The poster has clearly made an attempt by balancing the equation (albeit with a mistake - it should be 4CO2 on the right side) and is asking why the total number of moles is not conserved. – f'' Jul 22 '16 at 3:01
• While I can see the effort in the question, and hence will reopen it, I strongly recommend writing out the whole equation, i.e. $\ce{2HCCH + 5O2 -> 4CO2 + 2H2O}$. Then you can see, that acetylene is decomposing when combustion occurs. Since you make a dramatic change in the molecular geometry of the compounds, a volume change is certainly expected to occur, too. Also make sure that you understand that these are molar ratios and only within the ideal gas approximation they are proportional to the occupying volumes. – Martin - マーチン Jul 22 '16 at 8:33
• @Martin Thanks for enlightening. Is that the right hand side occupies less space? – Mathivanan Palraj Jul 22 '16 at 10:19

The volume can change in the course of the reaction. For example, if you add $\ce{Zn}$ to $\ce{HCl}$ the volume of the system changes. In the reaction $\ce{2H2 + O2 -> 2H2O}$ three volumes of gases turn into 1 (or even zero, if you allow water to condensate).
Behavior of gases (under low pressure) is described by ideal gas theory. In gas a molecule is a small particle that has a large free space around it. The amount of free space is independent from of the molecular mass. lightest molecule $\ce{H2}$ occupies as much space as a heavy gas $\ce{SF6}$. As a result, density of $c\ce{H2}$ is much smaller than density of air (this is how helium baloons fly) which in turn is smaller than density of $\ce{SF6}$.
From 12 carbons and 24 hydrogens you can make 6 molecules of ethane ($\ce{C2H4}$) or 4 molecules of cyclopropane ($\ce{C3H8}$) or three molecules of cyclobutane ($\ce{C4H8}$). The amount of atoms is the same, the weight is conserved, but the volume and density are not.