Question

(SOS) If 5.180 x 10ˆ24 atoms of water are made, how many moles of propane are needed? What about grams of oxygen?

The equation we use to help us here is C3H8 + 5 O2= 4 H2O + 3 CO2

We had just gone over conversion and I'm confused as to how to set it up to solve.
Any explanation is appreciated!

Answer 1

`C_3H_8(g) + 5O_2(g) rarr 3CO_2(g)+4H_2O(g) + Delta`

I make it a mass of approx. `10*g` propane....

Explanation:

A stoichiometric equation is VITAL and should be the first thing you write. And with these combustion reactions of hydrocarbons, the usual rigmarole is to balance the carbons as carbon dioxide, the hydrogens as water, and THEN balance the oxygens.

As written the equation is balanced....

With respect to water, we make a molar quantity of...

`(5.18xx10^23*"water molecules")/(6.022xx10^23*"water molecules"*mol^-1)=0.860*mol`

And CLEARLY ONE QUARTER of this molar quantity with respect to propane is combusted...i.e. `0.215*mol`. Do you follow? All I am doing is following the stoichiometry of the given equation.

And so .............

`"mass of propane"=0.215*molxx44.10*g*mol^-1=9.48*g`

I leave it to you to solve for the mass of dioxygen gas.