1) The air in a bicycle tire is bubbled through water and collected at 25°C. If
ID: 906151 • Letter: 1
Question
1) The air in a bicycle tire is bubbled through water and collected at 25°C. If the total volume of gas collected is 5.50 L at a temperature of 25°C and a pressure of 751 torr, how many moles of gas were in the bicycle tire?
________ mol
2) A particular deep-sea diving mixture contains 4.0 g of oxygen to every 96.0 g of helium. What is the partial pressure of oxygen when this mixture is delivered at a total pressure of8.1 atm?
_____atm
3) Consider the chemical reaction:
What mass of H2O is required to form 1.7 L of O2 at a temperature of 373 K and a pressure of 0.951 atm?
_____ g
4) Oxygen gas reacts with powdered aluminum according to the reaction:
What volume of O2 gas (in L), measured at 751 mmHg and 23°C, completely reacts with 50.0 g Al?
_____L
5) Lithium reacts with nitrogen gas according to the reaction:
What mass of lithium (in g) reacts completely with 52.6 mL of N2 gas at STP?
Explanation / Answer
1) according to PV = nRT
P = 751 torr = 0.988 atm.
V = 5.5 lt
R = 0.0821 Lt atm mol-1K-1
T = 298K
n = PV/RT
n = (0.988*5.5)/(0.0821*298)
n = 0.22 moles
2) partial pressure of O2 = molefraction of O2*(totalpressure of gas)
molefraction of O2 = (4/32)/[(4/32) + (96/4)]
= 0.005
partial pressure of O2 = 0.005*8.1 = 0.042 atm.
3) mass = P*V*mol.wt/(RT)
mass = (0.951*1.7*18)/(0.0821*373) = 0.95 gm.
4) V = nRT/P
P = 751 mmhg = 0.988 atm.
V = (50/27)*(0.0821*296)/0.988
V = 45.54 lt.
5)
n = PV/RT
n = (1*0.0526)/(0.0821*273)
n = 0.0023 moles
6 Li(s) + N2(g)......>2 Li3N(s)
according to given reaction 1mole of N2 reacts with 6 moles of Li
0.0023 moles of N2 reacts ...............?
= 0.0141 mole
mass of Li = no .of moles of Li* atomic.wt of Li
= 0.0141*6 = 0.0846 gm.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.