2C3H8 + 702 6 CO + 8 H0 If 2.50g of the organic fuel was reacted with 2.50g of oxygen and 1.50g of water was actually produced, what is the percent yield of this reaction? (stoichiometry needed was done for you, so all you have to do is calculate percent yield) Round to 3 significant figures and include a percent sign 2.50qGB Imol (Hg 8 mol H0 44.09g (sh 2 mol (3Hs 2.5090 Imole 02 8 mol H0 32.009 0 7mol 0 I 18.02gt0 = 4.09g HO Imol H0 18.029 420 = 1.61 g H0 Imol HO 1 pts