Ammonia is formed by the reaction 3H2 + N2 ---> NH3 (ammonia)?

Question

a. How many liters of ammonia at STP are formed when 25 L of N2 at STP react with H2.


b. How many liters of H2 AT 25 Centigrade and 1 atm are needed to react with 398L of N(2) Centigrade and 1 atm.


c.How many liters of NH3 at STP are formed when 9.56L of N2 at STP reacts with 27.1 L of H2 AT stp (This is a limiting reactant problem)



This problem gave me the fluxxxx. I worked it through though...Let's see if you can.

Answer 1

The balanced chem. equation is:

3H2(g) + N2(g) --> 2NH3(g)

a.

1 L of N2 gives 2 L of NH3 at stp
25 L of N2 gives 2*25 = 50 L of NH3 measured at stp.

b. I suppose that N2 is also at 25 oC.

1 L of N2 reacts with 3 L of H2
398 L of N2 reacts with 3*398 = 1194 L of H2.

c.

1 L of N2 reacts with 3 L of H2
9.56 L of N2 reacts with 3*9.56 = 28.68 L H2

But we have 27.1 L < 28.68 L of H2. So H2 is the limiting reagent ant reacts fully:

3 L of H2 gives 2 L of NH3
27.1 L of H2 gives 2*27.1/3 = 18.67 L of NH3

Answer 2

Let's see if we can...haha! Nice disguise

but let's have a shot

PV=nRT

1. STP=273 K, 1 atm

1(25)=n(.08206)(273)

1.12 mole N2=n

now 1.12 mole N2 * (1mole ammonia/1 mole N2)

1.12 mole NH3 formed

back to PV=nRT

So you since you have the same moles, same temp, and same pressure, it will be the same....25 L

2. (1atm)(398L)=n(.08206)(25+273=298K)
n=16.28 mol N2

16.28 N2* (3 mole H2)/(1 mole N2)= 48.84 mole H2

plug again

(1 atm)V=(48.84)(.08206)(298)
V= 1194.33 L H2

3. okay go through the PV=nRT to find how many moles of each you have with the given liters and STP

you have .427 mol N2 and 1.21 mol H2

Now find out how much of each you need

.427 mol N2 *(3 mole H2/1 mol N2)=1.28 mol H2 needed
1.21 mol H2 * (1/3)=.40 mol N2 needed

You dont have enough H2, so it is the limiting reactant...use it

So.......1.21 mol H2 *(1 mol NH3/3 mole H2)=.40 mol NH3

and plug

(1)V=(.40)(.08206)(273)= 8.96 L NH3 formed


jeez...i better get best answer for all that