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Before Bell Rings Chill out.

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Presentation on theme: "Before Bell Rings Chill out."— Presentation transcript:

1 Before Bell Rings Chill out.

2 Plan for Today Entry Task Popcorn Lab
Ideal Gas Law with Stoic and Limiting Reactants Study/Work Time

3 Moles + Stoic Test If you need to take/complete the test from last unit, come by this week during PH, before, or after school.

4 Moles + Stoic Test Corrections
Come do corrections thus week during PH, before, or after school Rework the problems you missed, with my help and your notes Once you have all the questions you missed corrected, score increased to a 70%. Don’t have to retake.

5 ETD 45 What is the pressure of 1.50g of helium at – 50.0C in a rigid container whose volume is 25.0 liters?

6 ETD 45 ANS What is the pressure of 1.50g of helium at – 50.0C in a rigid container whose volume is 25.0 liters?

7 Learning Targets I can determine what I need to study for the test next class.

8 Then find Water Vapor Pressure to Pop the kernel? How?
PV = nRT

9 How much pressure does it take to pop a kernel?
Calculations: Did well if in this range Water Vapor Pressure to Pop Kernel = 200 to 300 atm How many psi’s?

10 Stoichiometry + Ideal Gas Law
Sodium Azide (NaN3): Explosive compound in airbags Accident it decomposes Equation:  2 NaN3 (s)  2 Na (s) N2 (g)

11 #1 2 NaN3 (s)  2 Na (s) N2 (g) #1. If an airbag contains 7.5 g of NaN3, what volume of nitrogen gas will be produced at 25 °C and 1.2 atm? 7.5 g x (1 mole/65 g) = moles NaN3 x (3 moles N2 x 2 moles NaN3) = moles N2 PV = nRT 1.2 atm (V) = moles (0.082) (298 K) V = 3.52 L PV=nRT 0.98 atm (45.300L) = n (0.082)(293 K) n= 1.85 moles N2 1.85 moles N2 x (2 moles NaN3 / 3 moles N2) x (65 g NaN2 / 1 mole NaN3) = 80.2 g

12 #1 Ans 2 NaN3 (s)  2 Na (s) + 3 N2 (g)
If an airbag contains 7.5 g of NaN3, what volume of nitrogen gas will be produced at 25 °C and 1.2 atm? 7.5 g x (1 mole/65 g) = moles NaN3 x (3 moles N2 x 2 moles NaN3) = moles N2 PV = nRT 1.2 atm (V) = moles (0.082) (298 K) V = 3.52 L PV=nRT 0.98 atm (45.300L) = n (0.082)(293 K) n= 1.85 moles N2 1.85 moles N2 x (2 moles NaN3 / 3 moles N2) x (65 g NaN2 / 1 mole NaN3) = 80.2 g

13 #2 2 NaN3 (s)  2 Na (s) N2 (g) #2. If an airbag needs to inflate to a volume of 45300 mL, what mass of sodium azide is required at 20. °C and 0.98 atm? 7.5 g x (1 mole/65 g) = moles NaN3 x (3 moles N2 x 2 moles NaN3) = moles N2 PV = nRT 1.2 atm (V) = moles (0.082) (298 K) V = 3.52 L PV=nRT 0.98 atm (45.300L) = n (0.082)(293 K) n= 1.85 moles N2 1.85 moles N2 x (2 moles NaN3 / 3 moles N2) x (65 g NaN2 / 1 mole NaN3) = 80.2 g

14 #2 ANS 2 NaN3 (s)  2 Na (s) + 3 N2 (g)
#2. If an airbag needs to inflate to a volume of 45300 mL, what mass of sodium azide is required at 20. °C and 0.98 atm? 7.5 g x (1 mole/65 g) = moles NaN3 x (3 moles N2 x 2 moles NaN3) = moles N2 PV = nRT 1.2 atm (V) = moles (0.082) (298 K) V = 3.52 L PV=nRT 0.98 atm (45.300L) = n (0.082)(293 K) n= 1.85 moles N2 1.85 moles N2 x (2 moles NaN3 / 3 moles N2) x (65 g NaN2 / 1 mole NaN3) = 80.2 g

15 Limiting Reactants with Ideal Gas
NH3(g) + O2(g)  N2(g) + H2O(g) If 6.0 grams of ammonia react with 9.0 grams of oxygen, how many liters of nitrogen are produced if it is collected at 28C and a pressure of 3.5 atm?

16 NH3(g) + O2(g)  N2(g) + H2O(g)
If 6.0 grams of ammonia react with 9.0 grams of oxygen, how many liters of nitrogen are produced if it is collected at 28˚C and a pressure of 3.5 atm?

17 Ammonia gas (NH3) combines with oxygen gas (O2) to form nitrogen gas and water vapor. If 6.0 grams of ammonia react with 9 grams of oxygen, how many liters of nitrogen are produced if it is collected at 28C and a pressure of 3.5 atm?

18 Learning Targets I can determine what I need to study for the test next class.

19 Homework Pop Corn Lab- Extra Credit
Due next class Review sheet won’t be checked, just to help study for test. Test next class.


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