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Gas Stoichiometry

Gas Stoichiometry. Example 1. What volume of boron bromide at 300°C and 65 kPa would produce 600 L of hydrogen bromide? Given the following reaction: 2BBr 3 + 3H 2  2B + 6HBr. Givens: T = 300°C  273 + 300 = 573 K P = 65 kPa R = 8.31 ( kPa·L / mol·K ) V HBr = 600 L V BBr3 = ?

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Gas Stoichiometry

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  1. Gas Stoichiometry

  2. Example 1 What volume of boron bromide at 300°C and 65 kPa would produce 600 L of hydrogen bromide? Given the following reaction: 2BBr3 + 3H2 2B + 6HBr

  3. Givens: T = 300°C  273 + 300 = 573 K P = 65 kPa R = 8.31 (kPa·L/mol·K) VHBr = 600 L VBBr3 = ? • Using mole ratio, calculate VBBr3: VBBr3 = VHBr x ___________ = 600 L x 2 mol BBr3 6 mol HBr = 200 L If we used the ideal gas law, the missing variable we have here is n (# of moles). However, we cannot calculate for this using the equation as we do not have the right information, but.... mol BBr3 2 6 mol HBr

  4. Example 2 Oxygen gas can be prepared by decomposing potassium nitrate, as shown below: 2KNO3 (s)  2KNO2 (s) + O2 (g) What volume of O2 at 35°C and 1.18 atm can be produced from the decomposition of 35.5 g KNO3?

  5. Givens: T = 35°C  273+35 = 308 K P = 1.18 atm R = 0.0821 (atm·L/mol·K) mKNO3 = 35.5 g • Find # moles given (KNO3) nKNO3 = m/M = 35.5g/101.11 g/mol = 0.351 mol

  6. mol O2 1 2 mol KNO3 2. Find # moles required (O2) using mole ratio nO2 = nKNO3 x _________ = 0.351 mol x 1 mol O2 2 mol KNO3 = 0.176 mol 3. Find VO2 using ideal gas law eq’n PV = nRT (1.18 atm)V = (0.176 mol)(0.0821atm·L/mol·K)(308 K) V = 3.77 L

  7. Example 3 CH4 burns in O2, producing CO2(g)and H2O(g). A 1.22 L CH4 cylinder, at 15°C, registers a pressure of 328 kPa. How many grams of water vapour are produced? CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Givens: V = 1.22 L T = 15C  273 +15 = 288 K P = 328 kPa R = 8.31 (kPa·L/mol·K)

  8. Find # moles given (CH4) using ideal gas law eq’n PV = nRT (328 kPa)(1.22 L) = n(8.31kPa·L/mol·K)(288 K) n = 0.167 mol 2. Find # moles of required (H2O) using mole ratio nH2O = nCH4 x ___________ = 0.167 mol x 2 mol 1 mol = 0.334 mol 2 mol H2O 1 mol CH4

  9. 3. Find mass of water m = n x M = 0.334 mol x 18.02 g/mol =6.02 g

  10. Unit 4 Quest Tues. July 24th (10-11:30)Note – when you are done test, you will work QUIETLY on exam review or prep for summative • 10 K/U, 16 T/I, and 8 A • 10 MC (half are theoretical. Make sure you know KMT, the laws and theoretical scenarios. • Eg. Temperature of a gas remains constant while the volume of a given amount of gas doubles. What is the final volume? • Short calculations (4) • 1 Stoichiometry question • 3 gas applications (use your knowledge of KMT and gas laws)

  11. Summative – Weds. July 25th • Topic: Titration • Use basis of 2nd formal lab to help you with this • No notes allowed. CLOSED BOOK!You will be given only a procedure sheet and a handout to fill in answers • No talking amongst each other during lab!!!! • 14 before the break, and 14 after the break • You MUST complete titration in your allocated time! No extra time will be given due to equipment allocation. • 35 marks total (marks are also included for safety, competence, clean-up and quality/presentation of work in addition to calculations • So what will I do once the summative is handed in? You will work on exam review QUIETLY and INDEPENDENTLY.

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