Gas Stoichiometry Solutions
Boyle's Law
1. (2.3)(3.2) = 2 (P)
P = 3.68 atm
2. The graph rises by 0.02 every horizontal step of 0.01, so the slope, or k, is 2.
Thus (6.2)(V) = 2, and V = .323 L
3. Let x = original Volume
Let y = final Volume
(x)(2.5 atm) = y(2.0)
5x=4y Ratio = 4:5
4. (5)(V) = 3.5(4)
V = 2.8 L
P = 3.68 atm
2. The graph rises by 0.02 every horizontal step of 0.01, so the slope, or k, is 2.
Thus (6.2)(V) = 2, and V = .323 L
3. Let x = original Volume
Let y = final Volume
(x)(2.5 atm) = y(2.0)
5x=4y Ratio = 4:5
4. (5)(V) = 3.5(4)
V = 2.8 L
Charles's Law
1. (100mL)/(21°C) = (20mL)/(x)
(100mL)/(294°K) = (20mL)/(x)
x= 58.8°K
2. 2L = x L x = 2.88L
21°C 150°C
3. 0.5 L/21°C = x L/100°C 0.5 L/294°K = x L/373°K x= 0.634 L
Will not explode
4. 0.75L/293°K = x L/373°K x=0.64L
(100mL)/(294°K) = (20mL)/(x)
x= 58.8°K
2. 2L = x L x = 2.88L
21°C 150°C
3. 0.5 L/21°C = x L/100°C 0.5 L/294°K = x L/373°K x= 0.634 L
Will not explode
4. 0.75L/293°K = x L/373°K x=0.64L
Gay - Lussac's Law
1. 50°F = 10°C = 283°K 77°F = 25°C = 298°K
650 torr/283°K = 1.1 atm/298°K
650 torr/283°K = x torr/298°K x= 684 torr therefore it will not pop.
2. 675mmHg = 0.888atm0.888 atm/288°K = x atm/313°K
3. 1700.0 torr/150°C = x torr/450°C 1700.0 torr/423°K = x torr/723°K x = 2905.7 torr 5% margin: (2760.4 -- 3051.0)
4. 1 -- 7 seconds = 5L -- 35L
5L/x°K = 35L/673°K
x = 96.14°K
(673 - 96.14)/7 = 82.41°K/second
650 torr/283°K = 1.1 atm/298°K
650 torr/283°K = x torr/298°K x= 684 torr therefore it will not pop.
2. 675mmHg = 0.888atm0.888 atm/288°K = x atm/313°K
3. 1700.0 torr/150°C = x torr/450°C 1700.0 torr/423°K = x torr/723°K x = 2905.7 torr 5% margin: (2760.4 -- 3051.0)
4. 1 -- 7 seconds = 5L -- 35L
5L/x°K = 35L/673°K
x = 96.14°K
(673 - 96.14)/7 = 82.41°K/second
Avogadro's Law
1. 35.6 grams XeF4 = 0.172 moles XeF4
0.172 moles XeF4 x 22.4L/mole = 3.85L
2. 40.0g CO2 = 0.909 moles CO2
250 mL/0.909 moles = 90mL/x moles
x = 0.327 moles x 44g/mol
= 14.4 grams = 0.0144kg
3. 213 L/100 moles = k1 = 2.13
115 L/ 80 moles = k2 = 1.44
k1 > k2 therefore the new chamber is not able to lift the human up.
4. total moles = 1 mole He + 0.5 mole Chlorine gas = 1.5 moles gas
150 mL/1.5 moles gas = x mL/ 1 mole gas
x = 100mL
1.6 gram He = 0.4 mole He
1 mole gas - 0.4 mole He = 0.6 mole Chlorine gas
0.6 moles Chlorine gas = 21.27 grams Chlorine gas.
2. 40.0g CO2 = 0.909 moles CO2
250 mL/0.909 moles = 90mL/x moles
x = 0.327 moles x 44g/mol
= 14.4 grams = 0.0144kg
3. 213 L/100 moles = k1 = 2.13
115 L/ 80 moles = k2 = 1.44
k1 > k2 therefore the new chamber is not able to lift the human up.
4. total moles = 1 mole He + 0.5 mole Chlorine gas = 1.5 moles gas
150 mL/1.5 moles gas = x mL/ 1 mole gas
x = 100mL
1.6 gram He = 0.4 mole He
1 mole gas - 0.4 mole He = 0.6 mole Chlorine gas
0.6 moles Chlorine gas = 21.27 grams Chlorine gas.
Dalton's Law
1. P(total) = P(nitrogen) + P(oxygen) + P(carbon dioxide) + P(Helium) = 5 + 2 + 3 + 1 = 11atm
2. P(Giggity Gas) = P(Total) – P(Butterbeer) = 867torr – 69torr = 798torr
3. P(oxygen) = 0.3(5) = 1.5atm
4.
2. P(Giggity Gas) = P(Total) – P(Butterbeer) = 867torr – 69torr = 798torr
3. P(oxygen) = 0.3(5) = 1.5atm
4.
- P(carbon dioxide) = P(total) – P(water) = 749mmHg – 31.8mmHg = 717mmHg
- V = 0.400L
- T = 30 + 273 = 303K
- R = 0.0821 L x atm/(mol x K)
- PV = nRT
- n = (PV)/(RT)
- n = (717/760)(0.4)/((0.0821)(303)) = 0.0152 mol carbon dioxide
- 0.0152mol x (44.01g/mol) = 0.669 mol carbon dioxide
Ideal Gas Law
1. Sulfur Dioxide is limiting
0.50 mol SO2 = 0.50 mol SO3
0.50 mol SO3 x 22.4L/mol = 11.2L
2. (5 / 9) (9001°F - 32) = 4982.78°C
4982.78°C + 273 = 5255.78 K
n = 745 / (238.08 + 6* 18.998) = 2.081 moles
P(0.75) = 2.081 * 0.0821 * 5255.78
P = 1197.04 atm
The ring will be destroyed
3. Since all variables besides temperature and pressure are kept constant, this becomes a Gay - Lussac's Law problem, where k = 0.0821.
1.6 = 0.0821 (T1) and 1.2 = 0.0821 (T2)
The change in temperature is thus - 4.87 K = -4.87°C. Zhang's ramen gets pretty cold :(
4. (2.3)(1337) = n (0.0821)(564)
n = 66.411 moles of gas.
66.411 / 10 = 6.641 seconds.
Singed can continue to release gas for 6.6 seconds
0.50 mol SO2 = 0.50 mol SO3
0.50 mol SO3 x 22.4L/mol = 11.2L
2. (5 / 9) (9001°F - 32) = 4982.78°C
4982.78°C + 273 = 5255.78 K
n = 745 / (238.08 + 6* 18.998) = 2.081 moles
P(0.75) = 2.081 * 0.0821 * 5255.78
P = 1197.04 atm
The ring will be destroyed
3. Since all variables besides temperature and pressure are kept constant, this becomes a Gay - Lussac's Law problem, where k = 0.0821.
1.6 = 0.0821 (T1) and 1.2 = 0.0821 (T2)
The change in temperature is thus - 4.87 K = -4.87°C. Zhang's ramen gets pretty cold :(
4. (2.3)(1337) = n (0.0821)(564)
n = 66.411 moles of gas.
66.411 / 10 = 6.641 seconds.
Singed can continue to release gas for 6.6 seconds