Question

If `0.55*mol` of dioxygen react with excess dihydrogen, what mass of water results?

Answer 1

Look at the stoichiometry of the equation that gives water....

Explanation:

We could write....

`underbrace(H_2(g))_(2*g) + underbrace(1/2O_2(g))_(16*g) rarr underbrace(H_2O(l))_(18*g)`

....else we could write.....

`underbrace(2H_2(g))_("2 moles") + underbrace(O_2(g))_("1 mole") rarr underbrace(2H_2O(l))_("2 moles")`

In either case the stoichiometric equvialence is clear....

And if `0.55*mol` dioxygen react, we GETS a molar quantity of `0.55/2*mol` of water, i.e. a mass of `0.55/2*molxx18.01*g*mol^-1=4.95*g`