# How would you balance the complete combustion of acetylene by oxygen gas in a constant-pressure calorimeter?

##### 2 Answers

The basic equation is

The equation should be balanced as per the law of conservation of matter.

To balance the number of Hydrogen atoms on both the sides of the equation , I will add a

To balance the number of Carbon atoms on both the sides of the equation , I will add a

2

To balance the number of Oxygen atoms on both the sides of the equation , I will add 4 molecules of

2

Here's how I would approach it. To start off, the unbalanced equation was:

#"C"_2"H"_2(g) + "O"_2(g) -> "CO"_2(g) + "H"_2"O"(g)#

The hydrogens are balanced to begin with, so instinctively I would first balance the carbons by doubling the quantity of

#"C"_2"H"_2(g) + "O"_2(g) -> color(red)(2)"CO"_2(g) + "H"_2"O"(g)#

That changes the number of oxygens on the products side so that it is

(Fractional stoichiometries are OK, but are more conventionally used in enthalpy of formation reactions.)

So, since there are two oxygen atoms in one

#"C"_2"H"_2(g) + color(red)(5/2)"O"_2(g) -> color(red)(2)"CO"_2(g) + "H"_2"O"(g)#

You can verify that there are these numbers of atoms on each side:

#stackrel("Reactants Side")overbrace(2xx"C")# vs.#stackrel("Products Side")overbrace(2xx"C")#

#stackrel("Reactants Side")overbrace(5/2xx2xx"O")# vs.#stackrel("Products Side")overbrace(2xx2xx"O" + 1xx"O")#

#stackrel("Reactants Side")overbrace(2xx"H")# vs.#stackrel("Products Side")overbrace(2xx"H")#

So, technically, we balanced the equation. Now, just double everything to get rid of the fraction.

#color(blue)(2"C"_2"H"_2(g) + 5"O"_2(g) -> 4"CO"_2(g) + 2"H"_2"O"(g))#