I figured out how to do mole to mole and mass to mass but I am just lost on gas stoichiometry. Can someone please walk me through this problem? Thanks so much.

If 3.0 liters of N2 is combined with 8.0 liters of H2 at STP, according to

the equation below, how many liters of NH3 will be produced?

N2 + 3H2 --> 2NH3

## Answers (1)

Avogadro's Principle

equal volumes of gases at the same temp and pressusre have equal numbers of molecules

in other words

the volume occupied by a gas at certain cnditions is not dependent on what the molecules are bu only dependent on how mny molecules there are

now

numbers of molecules are usually measured in moles (6.022E23 molecules)

so it turns out that at STP there are 22.4 L/mol of a gas (providing that it acts close to ideally)

and

when that information is put into the combined gas law

V1P1/T1 = V2P2/T2

we get

PV = nRT

in the particular problem you have

N2 + 3H2 --> 2NH3

we could convert L to moles at STP

and proceed as usual

but

since T and P are constant, we can just use L

3.0. . 8.0

N2 + 3H2 --> 2NH3

now we have a limiting reactant problem

so

what if all 3.0L of N2 were used, then we'd need 9.0L of H2 (1 to 3 mole ratio)

but we only have 8.0 so H2 limits

and

. . . . .8.0. . . . 5.3L

N2 + 3H2 --> 2NH3

. . . . . .|_3 to 2_|

8.0/X = 3/2

X = 5.3 L