Question

When 1.57 mol `O_2` reacts with `H_2` to form `H_2O`, how many moles of `H_2` are consumed in the process?

Answer 1

Look at the stoichiometry, the chemical equivalence, of the reaction:

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

Explanation:

`H_2(g) + 1/2O_2(g) rarr H_2O(l)`, or I could simply double the equation,

`2H_2(g) + O_2(g) rarr 2H_2O(l)`

Now, clearly, for each mole/equiv of dioxygen, 2 moles/equivs dihydrogen react. If there are `1.57*mol` `O_2`, stoichiometric equivalence requires `2xx1.57*mol` dihydrogen. (i) How many moles of water will be generated; and (ii) what is the mass of the water?