1. Gas Stoichiometry A balanced equation shows the ratio of moles being used and produced Because of Avogrado’s principle, it also shows the ratio of volumes being used and produced

2. Example 2C 4 H 10(g) + 13O 2(g)  8CO 2(g) + 10 H 2 O (g) Shows molar and volume ratios Questions: –How many moles of oxygen are required to produce 35 moles of water? –How many liters of butane does it take to completely react with 50 liters of oxygen?

3. Try It What volume of oxygen (O 2 ) is needed to react with solid sulfur to form 3.5 L of SO 2 gas? O 2(g) + S (s)  SO 2 (g) so 3.5 L O 2 yields 3.5L SO 2 What volume of oxygen is needed to completely combust 2.36 L of methane gas? CH 4(g) + 2O 2 (g)  CO 2(g) + 2H 2 O (g) 2.36 L CH 4 X 2 mol O 2 / 1 mol CH 4 = 4.72 L O 2

4. Volume and Mass Problems A balanced equation indicates both moles and volume (for gases) and can convert either to mass Example: N 2(g) + 3H 2(g)  2NH 3(g) If 5.00 L of nitrogen reacts completely at a constant pressure and temperature of 3.00 atm and 298 K, how many grams of ammonia are produced?

5. Try It 1.Given: NH 4 NO 3(s)  N 2 0 (g) + 2H 2 O (g) Calculate the mass of solid ammonium nitrate that must be used to obtain.100 L of dinitrogen oxide gas at a pressure of 3.00 atm and a temperature of 15 o C. Hints: change.100 L of N 2 0 into moles calculate moles of NH 4 NO 3 that reacted calculate the mass of the NH 4 NO 3

6. Answers 1.NH 4 NO 3(s)  N 2 0 (g) + 2H 2 O (g) PV = nRT or n = PV / RT n = 3.00 atm x.100 L / 288 K x.0821 atm. L/K. mol =.0127 mol N 2 O .0127 mol NH 4 NO 3.0127 mol NH 4 NO 3 x 79.035 g/mol = 1.00 g NH 4 NO 3

7. Try It 2. Solid potassium metal will react with Cl 2 gas to form ionic potassium chloride. How many liters of Cl 2 gas are needed to completely react with.204 g of potassium at STP? 2K (s) + Cl 2(g)  2KCl (aq) Steps: convert grams of K into moles determine moles of Cl 2 needed convert moles of Cl 2 to liters

8. Answers 2K (s) + Cl 2(g)  2KCl (s).204 g K x 1 mol / 39.098 g =.00522 mol K.00522mol K x 1 mol Cl 2 / 2 mol K =.00261mol Cl 2 PV = nRT so V = nRT/P =.00261 mol Cl 2 x.0821 atm. L/K. mol x 273 K / 1.00 atm =.0585 L or 5.85 x 10 -2 L or 58.5 ml