C3H8 + 5O2 ---> 3CO2 + 4H2O
5 mols O2 for every mol of C3H8
a mol of O2 is 2*16 = 32 grams
a mol of C3H8 is 3*12+8 = 44 grams
so
11 grams of propane is 1/4 mol
so we need 5/4 mols of O2
(5/4)(32)
What mass of O2 would be needed to burn 11g of the propane gas?
6 answers
32g of oxygen and 44g of carbon ( iv) oxide each occupy 22.4 dm³ at S. T. P
James Daniel
Chemistry
How can I assist you with chemistry, James?
The correct answer is 5/4 x 32 = 40 grams O2 needed.