0.75 mole of solid `A_(4)` and 2 moles of gaseous `O_(2)` are heated in a sealed vessel, completely using up the reactants and producing only one compound. It is found that when the temperature is reduced to the initial temperature, the contents of the vessel exhibit a pressure equal to half the original pressure. What conclusions can be drawn from these data about the product of the reaction?

Explanation

First method: As both `A_(4)` and `O_(2)` are consumed
`("moles of" A_(4))/("moles of" O_(2)) =0.75/2.0`
`=(("moles of A")//4)/(("moles of O")//2) = 0.75/2.0`
`=("moles of A")/("moles of O") = 3/4`
Thus, the empirical formula of the product is `A_3O_4`. Further, as 2 moles of `O_(2)` give 1 mole of `A_3O_4` (for gases, pressure `prop` mole at constant temperature and volume), `A_3O_4` is also the molecular formula of the product.
Second Method : `underset("0.75 mole")(A_(4)s) + underset("2.0 mole")(O_(2) g) to underset("1.0 mole")(A_(x)O_(y)(g))`
Applying POAC for A atoms,
`4 xx` moles of `A_(4) = x xx `moles of `A_(x)O_(y)`
`4 xx 0.75 = x xx 1, x=3`
Applying POAC for O atoms.
`2 xx` moles of `O_(2)= y xx` moles of `A_(x)O_(y)`
`2 xx 2 = y xx 1, y =4`
In the following chapters, we shall apply the principle of atom conservation (POAC) along with the said rules in tackling the various problems encountered in chemical practice.