Question

The combustion of propane (C3H8) produces CO2 and H2O according to the following balanced equation: C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g) If the hydrocarbon is present in excess, what mass of oxygen (O2) in grams is necessary to form 12.9 g of CO2? Round your answer to the nearest 0.1.

Answer 1

CHx + 50, -7300, + 4H2O 191 19 9 19) from the above equation, a 3 mol of Co. required. : For Smol of quais => 1 mol or con requires =' mol Of Oz -O molor mass of Co, = 13.011 + Qx15.999) = 44.099 g/mol molar mass Vo = 2 x 15.999 - 31.998 g/mol 44. 1 from eqio ... 44 g of coa requares = 3 x 31.998 g of Oz Igor.co2 requires = 5 * 31.998 g of ₂ 1 = 1.212 g of O2 À 12.9 g of CO2 requires - 12.9 x1-312 = 15.6348 g of Oz. to form ..: 15.6348 g of Onygen will be required 12.9 g of co.