Consider the following chemical equation:
N2 + 3H2 = 2NH3
What volumes of nitrogen gas and hydrogen gas, each measured at 11 degrees Celsius and .998 atm, are needed to produce 5.00 g of ammonia?
I have been trying to use the ideal gas law but I don't know what to do with the 5.00 g of ammonia. Any help is appreciated. Thanks for your time :)!!
N2 + 3H2 = 2NH3
What volumes of nitrogen gas and hydrogen gas, each measured at 11 degrees Celsius and .998 atm, are needed to produce 5.00 g of ammonia?
I have been trying to use the ideal gas law but I don't know what to do with the 5.00 g of ammonia. Any help is appreciated. Thanks for your time :)!!
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You’ve written the reaction as if it might go to completion, but it’s actually an equilibrium reaction.
N2 + 3H2 <====> 2NH3
Regrettably, the rate of the forward reaction is extremely low at 11°C and 0.998 atm: you could put it all in a flask and wait for months, but you still wouldn’t find any appreciable amount of ammonia in the mixture.
(See http://en.wikipedia.org/wiki/Haber_proce…
Having said that, here’s how you solve the problem.
---------- ---------- ---------- ---------- ---------- ----------
The equation tells you that …
1 molecule N2 + 3 molecules H2 -----> 2 molecules NH3
1 mole N2 + 3 moles H2 -----> 2 moles NH3
Also, because all the species are gaseous, the equation tells you that ...
1 volume N2 + 3 volumes H2 -----> 2 volumes NH3
This gives you two possible approaches to answering the question: you can deal in moles, or you can deal in volumes. Since you’re asked to find volumes, we’ll go the volume route.
Step 1: Find the number of moles of NH3 formed.
You have to make 5.00 g NH3. Convert that to moles.
moles = mass / molar mass
= 5.00 g / 17.00 g/mol
= 0.294 mol
Step 2: Find the volume of NH3 produced at the specified temperature and pressure.
PV = nRT, so …
V = nRT / P
P = 0.998 atm = 101.1 kPa
n = 0.294 mol
R = 8.31 kPa*L / mol*K
T = 11°C = 284 K
V = (0.294 mol) (8.31 kPa*L / mol*K) (284 K) / 101.1 kPa
= 6.86 L
Step 3: Use the volume ratio to find the volumes of N2 and H2 under the same conditions of temperature and pressure.
Volume ratio N2 : H2 : NH3 = 1 : 3 : 2
Volume of NH3 = 6.86 L
Therefore,
volume of N2 = 3.43 L
volume of H2 = 10.3 L
---------- ---------- ---------- ---------- ---------- ----------
That’s it. I hope it’s clear how I arrived at this result, and that it’s of some help to you. Check my figures for errors. They look about right.
N2 + 3H2 <====> 2NH3
Regrettably, the rate of the forward reaction is extremely low at 11°C and 0.998 atm: you could put it all in a flask and wait for months, but you still wouldn’t find any appreciable amount of ammonia in the mixture.
(See http://en.wikipedia.org/wiki/Haber_proce…
Having said that, here’s how you solve the problem.
---------- ---------- ---------- ---------- ---------- ----------
The equation tells you that …
1 molecule N2 + 3 molecules H2 -----> 2 molecules NH3
1 mole N2 + 3 moles H2 -----> 2 moles NH3
Also, because all the species are gaseous, the equation tells you that ...
1 volume N2 + 3 volumes H2 -----> 2 volumes NH3
This gives you two possible approaches to answering the question: you can deal in moles, or you can deal in volumes. Since you’re asked to find volumes, we’ll go the volume route.
Step 1: Find the number of moles of NH3 formed.
You have to make 5.00 g NH3. Convert that to moles.
moles = mass / molar mass
= 5.00 g / 17.00 g/mol
= 0.294 mol
Step 2: Find the volume of NH3 produced at the specified temperature and pressure.
PV = nRT, so …
V = nRT / P
P = 0.998 atm = 101.1 kPa
n = 0.294 mol
R = 8.31 kPa*L / mol*K
T = 11°C = 284 K
V = (0.294 mol) (8.31 kPa*L / mol*K) (284 K) / 101.1 kPa
= 6.86 L
Step 3: Use the volume ratio to find the volumes of N2 and H2 under the same conditions of temperature and pressure.
Volume ratio N2 : H2 : NH3 = 1 : 3 : 2
Volume of NH3 = 6.86 L
Therefore,
volume of N2 = 3.43 L
volume of H2 = 10.3 L
---------- ---------- ---------- ---------- ---------- ----------
That’s it. I hope it’s clear how I arrived at this result, and that it’s of some help to you. Check my figures for errors. They look about right.