Question

When `10*g` dihydrogen gas, and `32*g` dioxygen gas react, which is the limiting reagent. And what mass of water will result?

Answer 1

You need (i) a stoichiometrically balanced equation...........

Explanation:

`H_2(g) + 1/2O_2(g) rarr H_2O(g)`

And (ii) equivalent quantities of dioxygen and dihydrogen.

`"Moles of dihydrogen"` `=` `(10*g)/(2.02*g*mol^-1)~=5*mol`.

`"Moles of dioxygen"` `=` `(32*g)/(32.0*g*mol^-1)~=1*mol`

Now, clearly, `O_2` is the limiting reagent, and there is a stoichiometric excess of `H_2`. At most `2 *mol` of `H_2` react quantitatively with the `1*mol` of `O_2`, to give `2*mol` of `H_2O`, which represents a mass of `36*g`.

At given temperature and pressure what volume of `"dioxygen"` will be required for a given volume of `"dihydrogen"`?