Question

What quantity of water results from the combustion of a `0.5*g` mass of dihydrogen gas?

Answer 1

Approx, `4.25*g`.

Explanation:

First, we need a stoichiometrically balanced equation, i.e.

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

`"Moles of dihydrogen"` `=` `(0.5*g)/(2xx1.00794*g*mol^-1)=0.25*mol`

Given the `1:1` stoichiometry, if `0.25*mol` dihydrogen react, then `0.25*mol` water result.

`"Mass of water"` `=` `0.25*cancel(mol)xx18.02*g*cancel(mol^-1)=??g`

How many grams of dioxygen were involved?