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Entry Task: April 18 th -19 th B.2

Entry Task: April 18 th -19 th B.2. Wrap up Proving “R” constant. We can find out the volume of gas through Stoichiometry. CH 4 + O 2  CO 2 + H 2 O. 2. 2. Think of the coefficients as volume ratios. 1 liter (volume) of CH 4. 2 liters (volume) of O 2.

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Entry Task: April 18 th -19 th B.2

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  1. Entry Task: April 18th-19th B.2 Wrap up Proving “R” constant

  2. We can find out the volume of gas through Stoichiometry CH4 + O2 CO2 + H2O 2 2 Think of the coefficients as volume ratios 1 liter (volume) of CH4 2 liters (volume) of O2 1 liter (volume) of CO2 2 liters (volume) of H2O

  3. CH4 + O2 CO2 + H2O 2 2 What if I have 3.5 liters of CH4, how much oxygen is needed for this reaction? 3.5 L of CH4 2 liters of O2 = 7 liters of O2 1 liter of CH4

  4. CH4 + O2 CO2 + H2O 2 2 If 4.00L of oxygen gas reacts completely by this reaction at a constant pressure and temperature of 2.00 atm and 300K, how many grams of water are produced? 4.00L O2 P = VO2 = T= • 2.00 atm 300K mH2O = R = 0.0821 X ALL data is based off of Oxygen gas

  5. CH4 + O2 CO2 + H2O 2 2 4.00L O2 P = VO2 = T= • 2.00 atm 300K mH2O = R = 0.0821 X Use the stoich- relationship between O2 and Water! 4.00 L of O2 2 liters of H2O = 4.00 liters of H2O 2 liter of O2

  6. CH4 + O2 CO2 + H2O 2 2 NOW we have a volume (4.00L of Water), next we can plug it into the ideal gas law. **noticed the change in volumes label!! P = VH2O = T= 4.00L H2O • 2.00 atm 300K mH2O = R = 0.0821 X

  7. PV = nRT P = VH2O = T= 4.00L H2O • 2.00 atm 300K mH2O = R = 0.0821 X (2.00 atm)(4.00 L) = (X H20)(0.0821)(300K)

  8. PV = nRT (2.00 atm)(4.00 L) = (X H20)(0.0821)(300K) (2.00 atm)(4.0 L) = X mol (0.0821)(300 K)

  9. DO the MATH (2.00 atm)(4.0 L) = X mol (0.0821)(300 K) 8 = 0.32 molH2O 24.63

  10. 0.32 molH2O Not done yet!!! We have to convert this to grams of H2O 0.32 molH2O 18 g of H2O = 5.85 g of H2O 1 of mol H2O

  11. Ammonia is synthesized from hydrogen and nitrogen gas. N2 + H2 NH3 3 2 If 5.00L of nitrogen reacts completely by this reaction at a constant pressure and temperature of 3.00 atm and 298K, how many grams of ammonia are produced?

  12. N2 + H2 NH3 3 2 If 5.00L of nitrogen reacts completely by this reaction at a constant pressure and temperature of 3.00 atm and 298K, how many grams of ammonia are produced? mNH3 = X R = 0.0821 VN2 = 5.00L P = 3.00 atm T = 298K ALL data is based off of Nitrogen gas

  13. N2 + H2 NH3 3 2 VN2 = 5.00L P = 3.00 atm T = 298K mNH3 = X R = 0.0821 Use the stoich- relationship between H2 and NH3! 5.0 L of N2 2 liters of NH3 = 10 liters of NH3 1 liter of N2

  14. NOW we have a volume (10L of NH3), next we can plug it into the ideal gas law. N2 + H2 NH3 3 2 VNH3 = 10 L P = 3.00 atm T1 = 298K mNH3 = X R = 0.0821

  15. PV = nRT mNH3 = X R= 0.0821 VNH3 = 10. L P = 3.00 atm T1 = 298K (3.00 atm)(10 L) = (X NH3)(0.0821 )(298K)

  16. PV = nRT (3.00 atm)(10 L) = (X NH3)(0.0821 )(298K) (3.00 atm)(10 L) = X mol (0.0821 ) (298 K)

  17. DO the MATH (3.00)(10 ) = X mol (0.0821 mol)(298) 30 = 1.23 mol NH3 24.47

  18. Are we done?? 1.23 mol NH3 No!! We have to convert this to grams of NH3 1.23 mol NH3 17.04 g of NH3 = 21.0 g of NH3 1 mol of NH3

  19. 1. Balance gas equation Summary on Gas Stoich problems 2. From balancing – get gas ratio 3. Use given volume to put into volume ratio to get unknowns’ volume 4. Plug numbers into ideal gas law equation 5. Get X by itself and do the math *6. If needed, convert mole to grams

  20. Ammonia nitrate is a common ingredient in a chemical fertilizer. Use the reaction shown to calculate the mass of solid ammonium nitrate that must be used to obtain 0.100 L of dinitrogenmonoxide gas at STP NH4NO3 N2O + H2O 2

  21. NH4NO3 N2O + H2O 2 There is 0.100 liters of N2O, how much oxygen is needed for this reaction 0.100 L of N2O 1 liters of NH4NO3 1 liter of N2O = 0.100 L of NH4NO3

  22. PV = nRT Ammonia nitrate is a common ingredient in a chemical fertilizer. Use the reaction shown to calculate the mass of solid ammonium nitrate that must be used to obtain 0.100 L of dinitrogen oxide gas at STP VNH4NO3= 0.100 L P = 1.00 atm T1 = 0.00 + 273 = 273K mNH4NO3 = X R= 0.0821 (1.00 atm)(0.100 LNH4NO3) =(XNH4NO3)(0.0821)(273K)

  23. PV = nRT (1.00 atm)(0.100 L) = (XNH4NO3)(0.0821)(273K) (1.00 atm)(0.100 L) = X molNH4NO3 (0.0821) (273K)

  24. DO the MATH (1.00)(0.100 ) = X molNH4NO3 (0.0821)(273) 0.1 22.41 = 0.00446 mol NH4NO3

  25. Are we done?? = 0.00446 mol NH4NO3 No!! We have to convert this to grams of NH4NO3 80.04 g of NH4NO3 0.00446 molNH4NO3 1 mol of NH4NO3 = 0.357 g of NH4NO3

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