Question

Is a `3.44*g` mass of oxygen gas sufficient to completely oxidize a `14.8*g` mass of propane?

Answer 1

`C_3H_8+5O_2rarr3CO_2 +4H_2O`

There are insufficient dioxygen for complete combustion......

Explanation:

And then it is simply a matter of working out the molar quantities:

`"Moles of propane"=(14.8*g)/(44.10*g*mol^-1)=0.336*mol`

`"Moles of dioxygen"=(3.44*g)/(32.00*g*mol^-1)=0.108*mol`

Oxygen is present in stoichiometric deficiency. We need further details of the experiment.