Question

What is the mass of oxygen required for the complete combustion of 16 oz of propane?

Answer 1

And so we gots `1*oz-=28.35*g`

Explanation:

We need (i) a stoichiometric equation...

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

And (ii) equivalent quantities of propane gas...

`n_"propane"=(16*ozxx28.35*g*oz^-1)/(44.10*g*mol^-1)=10.29*mol`

Clearly, we need `51.45*mol` `"dioxygen gas"`...for an equivalent quantity...

i.e. `51.45*molxx32.00*g*mol^-1~=1650*g` or if you like `58.1*oz`...and I would challenge any chemistry lab to still have a balance that used Imperial measures...

And for your next trick try the calculation of the speed of light in `"furlongs per fornight"`...