combustion of 1.7 L of this mixture? Assume ideal gas behavior and that products of the cobustion are CO2(g) and H2O(l)
2007-04-15 06:44:23 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Add together (0.78 x 1.7/22.4) x enthalpy of combustion of methane + (0.22 x 1.7/22.4) x enthalpy of combustion of ethane.
2007-04-16 06:41:39 · answer #1 · answered by Gervald F 7 · 0⤊ 0⤋
4 gm CH4 equals 4.0 / (sixteen.043) moles of CH4 or 0.249 moles. The balanced equation exhibits you that a million mole of CH4 will produce 2 moles of H2O. for this reason you will get 2 x 0.249 or 0.498 moles of H2O. as a results of fact this would be a combustion reaction the water would be interior the vapor (gas) state. the challenge tells you this as a results of fact it needed you to transform the quantity of H2O produced to an equivalent quantity of gas at STP. the ideal gas regulation says that a million mole of an ideal gas will occupy 22.4 liters at STP. for this reason 0.498 moles of gas will occupy 0.498 x 22.4 or 11.one hundred fifty five liters at STP.
2016-12-16 06:28:46 · answer #2 · answered by goslin 4 · 0⤊ 0⤋