1. Wrap up Proving “R” constant

2.  We can find out the volume of gas through Stoichiometry CH 4 + O 2  CO 2 + H 2 O22 Think of the coefficients as volume ratios 1 liter (volume) of CH 4 2 liters (volume) of O 2 1 liter (volume) of CO 2 2 liters (volume) of H 2 O

3. CH 4 + O 2  CO 2 + H 2 O22 What if I have 3.5 liters of CH 4, how much oxygen is needed for this reaction? 3.5 L of CH 4 1 liter of CH 4 2 liters of O 2 = 7 liters of O 2

4. CH 4 + O 2  CO 2 + H 2 O22 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? P = V O2 =T= m H2O = R = 2.00 atm 4.00L O 2 300K X 0.0821 ALL data is based off of Oxygen gas

5. CH 4 + O 2  CO 2 + H 2 O22 P = V O2 =T= m H2O = R = 2.00 atm 4.00L O 2 300K X 0.0821 Use the stoich- relationship between O 2 and Water! 4.00 L of O 2 2 liter of O 2 2 liters of H 2 O = 4.00 liters of H 2 O

6. CH 4 + O 2  CO 2 + H 2 O22 P = V H2O =T= m H2O = R = 2.00 atm 4.00L H 2 O 300K X 0.0821 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!!

7. PV = nRT (2.00 atm)(4.00 L) = (X H 2 0)(0.0821)(300K) P = V H2O =T= m H2O = R = 2.00 atm 4.00L H 2 O 300K X 0.0821

8. PV = nRT (2.00 atm)(4.0 L) (0.0821)(300 K) = X mol (2.00 atm)(4.00 L) = (X H 2 0)(0.0821)(300K)

9. DO the MATH 8 24.63 = 0.32 mol H 2 O (2.00 atm)(4.0 L) (0.0821)(300 K) = X mol

10. 0.32 mol H 2 O Not done yet!!! We have to convert this to grams of H 2 O 0.32 mol H 2 O 1 of mol H 2 O 18 g of H 2 O = 5.85 g of H 2 O

11. N 2 + H 2  NH 3 23 Ammonia is synthesized from hydrogen and nitrogen gas. 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. N 2 + H 2  NH 3 23 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? V N2 = 5.00L P = 3.00 atm T = 298K m NH3 = X R = 0.0821 ALL data is based off of Nitrogen gas

13. N 2 + H 2  NH 3 23 V N2 = 5.00L P = 3.00 atm T = 298K m NH3 = X R = 0.0821 5.0 L of N 2 1 liter of N 2 2 liters of NH 3 = 10 liters of NH 3 Use the stoich- relationship between H 2 and NH 3 !

14. N 2 + H 2  NH 3 23 NOW we have a volume (10L of NH 3 ), next we can plug it into the ideal gas law. V NH3 = 10 L P = 3.00 atm T 1 = 298K m NH3 = X R = 0.0821

15. PV = nRT V NH3 = 10. L P = 3.00 atm T 1 = 298K m NH3 = X R= 0.0821 (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) (0.0821 ) (298 K) = X mol

17. 30 24.47 = 1.23 mol NH 3 = X mol (3.00)(10 ) (0.0821 mol )(298)

18. 1.23 mol NH 3 No!! We have to convert this to grams of NH 3 1.23 mol NH 3 1 mol of NH 3 17.04 g of NH 3 = 21.0 g of NH 3

19. 1. Balance gas equation 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 dinitrogen monoxide gas at STP NH 4 NO 3  N 2 O + H 2 O2

21. There is 0.100 liters of N 2 O, how much oxygen is needed for this reaction 0.100 L of N 2 O 1 liter of N 2 O 1 liters of NH 4 NO 3 = 0.100 L of NH 4 NO 3 NH 4 NO 3  N 2 O + H 2 O2

22. PV = nRT V NH4NO3 = 0.100 L P = 1.00 atm T 1 = 0.00 + 273 = 273K m NH4NO3 = X R= 0.0821 (1.00 atm)(0.100 L NH4NO3 ) = (X NH4NO3 )(0.0821)(273K) 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

23. PV = nRT (1.00 atm)(0.100 L) = (X NH4NO3 )(0.0821)(273K) (1.00 atm)(0.100 L) (0.0821) (273K) = X mol NH4NO3

24. DO the MATH 0.1 22.41 = 0.00446 mol NH 4 NO 3 = X mol NH4NO3 (1.00)(0.100 ) (0.0821)(273)

25. = 0.00446 mol NH 4 NO 3 No!! We have to convert this to grams of NH 4 NO 3 0.00446 mol NH 4 NO 3 1 mol of NH 4 NO 3 80.04 g of NH 4 NO 3 = 0.357 g of NH 4 NO 3