Gas Law Stoichiometry PDF

Gas Law Stoichiometry  This is the same as any other stoichiometry 3 Steps - What do you think they are?  This is the same as any other stoichiometry – Convert to mole – Convert mole of known to mole of unknown – Convert to desired unit (with gas stoichiometry, a shortcut can...

Gas Law Stoichiometry  This is the same as any other stoichiometry 3 Steps - What do you think they are?  This is the same as any other stoichiometry – Convert to mole – Convert mole of known to mole of unknown – Convert to desired unit (with gas stoichiometry, a shortcut can sometimes be used) The ideal gas law just gives you another way to convert to mole amounts (under any condition, not just STP)  Summary of the last 3 units of SCH3U0 (ways to convert to/from moles): Using: Unit-Conversion Factor/Method: Particles Mass (of a pure substance): Volume (for solutions): Volume, Pressure & Temperature (for gases):  Summary of the last 3 units of SCH3U0 (ways to convert to/from moles): Using: Unit-Conversion Factor/Method: Avogadro’s number: (6.022x1023 particles) Particles 1mol Mass (of a pure substance): Molar mass (g/mol) Volume (for solutions): Concentration (mol/L) Volume, Pressure & PV = nRT Temperature (for gases):  Unit 3 particles Avogadro’ s number Unit 5 mol Unit 3 Gases PV=nRt Molar mass mass Concentration and Volume Solutions Unit 4 Eg. 1: Nitrogen and hydrogen gas react to form ammonia. If 1.00 L of nitrogen reacts with excess hydrogen gas how many liters of ammonia gas are produced under SATP? N + 3H → 2NH 2(g) 2(g) 3(g) Eg. 1: Nitrogen and hydrogen gas react to form ammonia. If 1.00 L of nitrogen reacts with excess hydrogen gas how many liters of ammonia gas are produced under SATP? N + 3H → 2NH 2(g) 2(g) 3(g) 2molNH3 0.0404molN2 x = NH3 1molN2 0.0808mol PV=nRT n= PV PV=nRT RT V= nRT n= (100kPa) (1.00L) P (8.314kPa L) (298K) V = (0.0808mol) (8.314kPa L/mol K)(298K) 100kPa mol K V = 2.00LNH3 n= 0.0404molN2 Therefore 2.00L of ammonia are produced Eg. 1: Nitrogen and hydrogen gas react to form ammonia. If 1.00 L of nitrogen reacts with excess hydrogen gas how many liters of ammonia gas are produced under SATP? (NOTE: With gasses, mols α volume (thank you Avogadro) so the volume ratio will follow the mol ratio) Eg. 1: Nitrogen and hydrogen gas react to form ammonia. If 1.00 L of nitrogen reacts with excess hydrogen gas how many liters of ammonia gas are produced under SATP? (NOTE: With gasses, mols α volume (thank you Avogadro) so the volume ratio will follow the mol ratio) N + 3H → 2NH 2(g) 2(g) 3(g) 2molNH3 1.00 LN2 x = NH3 1molN2 2.00L Therefore 2.00L of ammonia are produced (If conditions changed, we would have to use combined or ideal gas law) E.g. 3 What volume of hydrogen gas is produced when excess sulfuric acid reacts with 40.0 g of iron at 18.0 °C and 100.3 kPa? E.g. 3 What volume of hydrogen gas is produced when excess sulfuric acid reacts with 40.0 g of iron at 18.0 °C and 100.3 kPa? Solution mFe = 40.0 g T = 18.0 °C P = 100.3 kPa MFe = 55.85 g/mol Fe + H SO →H + (s) 2 4(aq) 2(g) FeSO 4(aq) V = 17.3 L  Pg. 603, Q. 1-5