Consider the unbalanced equation: C_6H_14 + O_2 -> CO_2 + H_2O. What mass of O_2 is required to react with "11.5 g" of C_6H_14?

1 Answer
Mar 15, 2017

"0.0400 g O"_2" is required to react with "11.5 g" of "C"_6"H"_14".

Explanation:

Start with a balanced equation.

"2C"_6"H"_14 + "19O"_2rarr"12CO"_2 + "14H"_2"O"

Use the balanced equation to determine the mole ratios between "C"_6"H"_14" and "O"_2".

("2 mol C"_6"H"_14)/("19 mol O"_2") and ("19 mol O"_2)/("2 mol C"_6"H"_14)

Determine the molar masses of "C"_6"H"_14 and "O"_2".

"C"_6"H"_14:"86.178 g/mol"
https://www.ncbi.nlm.nih.gov/pccompound?term=C6H14
"O"_2:"31.998 g/mol"

Determine the moles of "C"_6"H"_14" by dividing its given mass by its molar mass.

(11.5 cancel"g C"_6"H"_14)/(86.178cancel"g"/"mol")="0.13344 mol C"_6"H"_14"

Determine moles of "O"_2" by multiplying mol "C"_6"H"_14" by the mol ratio that has "O"_2" in the numerator.

0.13344cancel"mol C"_6"H"_14xx(19"mol O"_2)/(2 cancel"mol C"_6"H"_14)="1.2678 mol O"_2"

Determine the mass of "O"_2" that will react with 11.5"g C"_6"H"_14" by multiplying the moles "O"_2" by its molar mass.

1.2678cancel"mol O"_2xx(31.998"g O"_2)/(1cancel"mol O"_2)="0.0400 g O"_2" (rounded to three significant figures)