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

1 Answer
Sep 4, 2017

Answer:

#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.