1. How many grams of NH3, can we produce when 5.40 g of H2, is reacted with excess N2?
N2 + 3H2(g) → 2 NH3(g)

30.26 grams of N is produced when 5.40 g of , is reacted with excess nitrogen gas.

Explanation:

Data given:

mass of ammonia N =  ?

mass of Hydrogen gas = 5.40 grams

Balanced chemical reaction:

+ 3 ⇒2N

Atomic mass of hydrogen = 2.06 gram/mole

atomic mass of nitrogen = 28.006 grams/mole

atomic mass of ammonia = 17 grams/mole

moles of hydrogen provided

number of moles =

putting the values in the equation:

number of moles=

number of moles = 2.67 moles of hydrogen gas is given

from the chemical reaction:

3 moles of H2 gives 2 moles of NH3

SO, 2.67 moles of H2 will give x moles of NH3

=

3x = 5.34

x = 1.78 moles of ammonia

mass of ammonia = number of moles x atomic mass

mass of ammonia = 1.78 x 17

mass of ammonia = 30.26 grams

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g