Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperature. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1 g of NH
is reacted with 90.4 g of CuO, which is the limiting reagent? How many grams of N
Balance the equation
From the description above, we can translate the reaction into an equation as follows-
NH3 (g) + CuO (s) $\rightarrow$ Cu (s) + H2O (g) + N2 (g)
Balancing the above reaction we get:
2NH3 (g) + 3CuO (s) $\rightarrow$ 3Cu (s) + 3H2O (g) + N2 (g)
Compute the moles of the substances whose mass are given
Amount of NH3 given is 18.1g.
Amount of CuO given is 90.4 g.
1 mole of NH3 $\rightarrow$ 17.03 g of NH3
‘x’ moles of NH3 $\rightarrow$ 18.1 g of NH3
x = (18.1 g of NH3) (1 mole of NH3) / (17.03 g of NH3)
= 1.063 moles of NH3
1 mole of CuO $\rightarrow$ 79.54 g of CuO
‘y’ mole of CuO $\rightarrow$ 90.4 g of CuO
y = (90.4 g of CuO) (1 mole of CuO) / (79.54 g of CuO)
= 1.14 moles of CuO
To determine the limiting reagent, use the mole ratio. Compare the ratio of the moles obtained from the amount given and from the balanced equation.
The mole ratio of 2NH3: 3CuO = 2:3
2 moles of NH3 react with 3 moles of CuO
1.06 moles of NH3 $\rightarrow$ ‘a’ moles of CuO
a = (1.06 moles of NH3) (3 moles of CuO) / (2 moles of NH3)
= 1.59 moles of CuO
Thus, we require 1.59 moles of CuO to react with 1.06 moles of NH3. But we only have 1.14 moles of CuO. So, the amount of CuO is limiting.
So, the amount of the product formed will depend upon the amount of the limiting reactant.
CuO is the limiting reagent.
Use the amount or moles of the limiting reagent to determine the amount or moles of the unknown
The mole ratio of 3CuO: 1N2 = 3:1
3 moles of CuO $\rightarrow$ 1 mole of N2
1.14 moles of CuO $\rightarrow$ ‘b’ moles of N2
b = (1.14 moles of CuO) (1 mole of N2) / (3 moles of CuO)
= 0.38 moles of N2
1 mole of N2 $\rightarrow$ 28 g of N2
0.38 moles of N2 $\rightarrow$ ‘c’ g of N2
c = (0.38 moles of N2) (28 g of N2) / (1 mole of N2)
= 10.64 g of N2