## Kwa m11a1 nozzle

A student burns 40.0-g propane (C3H8) in oxygen to produce carbon dioxide and water at STP. How many liters of steam are produced? C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g)

10.0 g C3H8 X 1 mol C3H8/44.11 g C3H8 X 5 mol O2/1 mol C3H8 = 1.13 moles O2 = moles pure O2 needed. Since the ideal gas law says, PV = nRT

C3H8(l) + O2(g) --> CO2(g) + H2O(g) Which reaction will produce a greater volume of carbon dioxide at 22 oC and 0.998 atm if 500. g sugar and 500. g propane are used in each reaction with an excess of oxygen?

C3H8(g) + 4 Cl2(g) C3H4Cl4(g) + 4 HCl(g) A 6.0 mol sample of C3H8(g) and a 20. mol sample of Cl2(g) are placed in a previously evacuated vessel, where they react according to the equation above. After one of the reactants has been totally consumed, how many moles of HCl(g) have been produced? A) 4.0 mol B) 8.0 mol C) 20. mol D) 24 mol

C3H8 + 5O2 = 3CO2 + 4H2O We are given the amount of C3H8 to be reacted with O2. This will be the starting point of the calculations. 73.7 g C3H8 ( 1 mol C3H8/ 44.1 g C3H8 ) ( 5 mol O2 / 1 mol C3H8 ) ( 32.0 g O2 / 1 mol O2 ) = 267.39 g O2

CH4(g)+O2(g)→CO2(g)+H2O(g) The coefficients in a balanced equation, determine the ratio of the moles of reactants to the moles of products. The first step is to balance the equation.