gas-stoichiometry

Project in Chemistry : 

Project in Chemistry Maribeth M.Tamayo III-A Mr. Jerry Manansala Teacher

Gas Stoichiometry : 

Gas Stoichiometry We have looked at stoichiometry: 1) using masses & molar masses, & 2) concentrations. We can use stoichiometry for gas reactions. As before, we need to consider mole ratios when examining reactions quantitatively. At times you will be able to use 22.4 L/mol at STP and 24.8 L/mol at SATP as shortcuts. grams (x) ? moles (x) ? moles (y) ? grams (y)

Sample problem 1 : 

Sample problem 1 CH4 burns in O2, producing CO2 and H2O(g). A 1.22 L CH4 cylinder, at 15°C, registers a pressure of 328 kPa. What volume of O2 at SATP will be required to react completely with all of the CH4? First: CH4(g) + 2O2(g) ? CO2(g) + 2H2O(g) PV = nRT = n = 0.167 mol P = 328 kPa, V = 1.22 L, T = 288 K # mol O2= 0.167 mol CH4 = 0.334 mol PV = nRT =V = 8.28 L P= 100 kPa, n= 0.334 mol, T= 298 K or # L = 0.334 mol x 24.8 L/mol = 8.28 L

Sample problem 1 continued : 

Sample problem 1 continued CH4(g) + 2O2(g) ? CO2(g) + 2H2O(g) b) How many grams of H2O(g) are produced? c) What volume of CO2 (at STP) is produced if only 2.15 g of the CH4 was burned? # g H2O= 0.167 mol CH4 = 6.02 g H2O # mol CO2= 2.15 g CH4 = 0.134 mol CO2 PV = nRT P = 101.3 kPa, n = 0.134 mol, T = 273 K = V = 3.00 L CO2 or # L = 0.134 mol x 22.4 L/mol = 3.00 L

Sample problem 2 : 

Sample problem 2 Ammonia (NH3) gas can be synthesized from nitrogen gas + hydrogen gas. What volume of ammonia at 450 kPa and 80°C can be obtained from the complete reaction of 7.5 kg hydrogen? # mol NH3= 7500 g H2 = 2475 mol PV = nRT P = 450 kPa, n = 2475 mol, T = 353 K = V = 16 135 L NH3 First we need a balanced equation: N2(g) + 3H2(g) ? 2NH3(g)

Slide 7: 

Gas Stoichiometry

Sample problem 3 : 

Sample problem 3 Hydrogen gas (and NaOH) is produced when sodium metal is added to water. What mass of Na is needed to produce 20.0 L of H2 at STP? First we need a balanced equation: 2Na(s) + 2H2O(l) ? H2(g) + 2NaOH(aq) # g Na= 0.893 mol H2 = 41.1 g Na PV = nRT = n = 0.893 mol H2 P= 101.3 kPa, V= 20.0 L, T= 273 K or # mol = 20.0 L x 1 mol / 22.4 L = 0.893 mol

Assignment : 

Assignment What volume of oxygen at STP is needed to completely burn 15 g of methanol (CH3OH) in a fondue burner? (CO2 + H2O are products) When sodium chloride is heated to 800°C it can be electrolytically decomposed into Na metal & chlorine (Cl2) gas. What volume of chlorine gas is produced (at 800°C and 100 kPa) if 105 g of Na is also produced? What mass of propane (C3H8) can be burned using 100 L of air at SATP? Note: 1) air is 20% O2, so 100 L of air holds 20 L O2, 2) CO2 and H2O are the products of this reaction.

Slide 11: 

A 5.0 L tank holds 13 atm of propane (C3H8) at 10°C. What volume of O2 at 10°C & 103 kPa will be required to react with all of the propane? Nitroglycerin explodes according to: 4 C3H5(NO3)3(l)? 12 CO2(g) + 6 N2(g) + 10 H2O(g) + O2(g) a) Calculate the volume, at STP, of each product formed by the reaction of 100 g of C3H5(NO3)3. b) 200 g of C3H5(NO3)3 is ignited (and completely decomposes) in an otherwise empty 50 L gas cylinder. What will the pressure in the cylinder be if the temperature stabilizes at 220°C?

Answers : 

Answers 1. 3O2(g) + 2CH3OH(l) ? 2CO2(g) + 4H2O(g) # L O2= 15 g CH3OH = 15.7 L O2 2. 2NaCl(l) ?2Na(l) + Cl2(g) # mol Cl2= 105 g Na PV = nRT P = 100 kPa, n = 2.284 mol, T = 1073 K = V = 204 L Cl2 = 2.284 mol Cl2

Slide 13: 

3. C3H8(g) + 5O2(g) ? 3CO2(g) + 4H2O(g) # g C3H8= 20 L O2 = 7.1 g C3H8 4. C3H8(g) + 5O2(g) ? 3CO2(g) + 4H2O(g) PV = nRT # mol O2= 2.8 mol C3H8 = 14 mol O2 n = = 2.8 mol C3H8 PV = nRT P = 103 kPa, n = 14 mol, T = 283 K = V = 320 L O2

Slide 14: 

5. # mol C3H5(NO3)3= 100 g C3H5(NO3)3 = 0.4403 mol # L CO2= 0.4403 mol C3H5(NO3)3 = 29.6 L CO2 # L N2= 0.4403 mol C3H5(NO3)3 = 14.8 L N2 # L H2O= 0.4403 mol C3H5(NO3)3 = 24.7 L H2O # L O2= 0.4403 mol C3H5(NO3)3 = 2.47 L O2

Slide 15: 

5. # mol C3H5(NO3)3= 200 g C3H5(NO3)3 = 0.8806 mol # mol all gases= 0.8806 mol C3H5(NO3)3 = 6.385 mol all gases PV = nRT V = 50 L, n = 6.385 mol, T = 493 K = P = 523 kPa For more lessons, visit www.chalkbored.com