Question

What is the number of oxygen molecules needed to produce 3.232 g of water?

Would love to know HOW you found it! Thanks so much :)

Answer 1

We have to assess a stoichiometric reaction.....

`0.090*molxx6.022xx10^23*mol^-1=5.40xx10^22` with respect to DIOXYGEN MOLECULES.....

Explanation:

And thus....

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

And give stoichiometric equivalence this tells me that `2*g` of dihydrogen, i.e. one mole with respect to dihydrogen, reacts with half a mole of dioxygen gas, i.e. `16*g`, to give `18*g` of water.

Are you with me? Mass is conserved in all chemical reaction.

And so we calculate the molar quantity of water...

`"Moles of water"-=(3.232*g)/(18.01*g*mol^-1)=0.180*mol`

An given the state stoichiometry....there MUST HAVE BEEN a molar quantity of `0.090*mol` dioxygen reactant...with me?

And thus the number of dioxygen molecules....

`0.090*molxx6.022xx10^23*mol^-1=5.40xx10^22` with respect to DIOXYGEN.....

`"Pretty vast but not infinite....so we hadn't engaged the "`
`"infinite improbability drive"`