Nitrogen dioxide reacts with water to produce oxygen and ammonia: 4NO2(g)+6H2O(g
ID: 707223 • Letter: N
Question
Nitrogen dioxide reacts with water to produce oxygen and ammonia: 4NO2(g)+6H2O(g)?7O2(g)+4NH3(g)
How many grams of NH3 can be produced when 3.80 L of NO2 reacts at 415 ?C and 735 mmHg ?
Express your answer with the appropriate units.
A sample of methane (CH4) has a volume of 26 mL at a pressure of 0.87 atm . What is the final volume, in milliliters, of the gas at each of the following pressures, if there is no change in temperature and amount of gas?
0.20 atm
2.30 atm
2500 mmHg
65.0 torr
Express your answer using two significant figures!!!!
Explanation / Answer
4NO2(g)+6H2O(g)-------------->7O2(g)+4NH3(g)
P = 735mmHg
= 735/760 = 0.967 atm
V = 3.8L
T = 415+273 = 688K
PV = nRT
n = PV/RT
= 0.967*3.8/0.0821*688 = 0.065moles
4NO2(g)+6H2O(g)-------------->7O2(g)+4NH3(g)
4 moles of NO2 react with H2O to gives 4 moles of NH3
0.065 moles of NO2 react with H2O to gives = 4*0.065/4 = 0.065 moles of NH3
mass of NH3 = no of moles * gram molar mass
= 0.065*17 = 1.105g
initial Final
P1 = 0.87atm P2 = 0.2atm
V1 = 26ml V2 =
P1V1 = P2V2
V2 = P1V1/P2
= 0.87*26/0.2 = 113ml
initial Final
P1 = 0.87atm P2 = 2.3atm
V1 = 26ml V2 =
P1V1 = P2V2
V2 = P1V1/P2
= 0.87*26/2.3 =9.8ml
nitial Final
P1 = 0.87atm P2 = 2500mmHg = 2500/760 = 3.29atm
V1 = 26ml V2 =
P1V1 = P2V2
V2 = P1V1/P2
= 0.87*26/3.29 6.9ml
nitial Final
P1 = 0.87atm P2 = 65torr = 65/760 = 0.0855atm
V1 = 26ml V2 =
P1V1 = P2V2
V2 = P1V1/P2
= 0.87*26/0.0855 = 264.6ml
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