Question

Please help, its urgent.

165.0 grams of nitrogen gas reacts with 215.0 grams of hydrogen gas: N2 + 3H2--> 2NH3.
a) Identify the limiting reagent:
b) Calculate the grams of ammonia formed:
c) Calculate the grams of excess reactant formed:

Answer 1

Mass of excess reactant left = 179.6 g

Limiting reactant = nitrogen

Mass of ammonia formed = 200.6 g

Step-by-step explanation:

Given data:

Mass of nitrogen = 165.0 g

Mass of hydrogen = 215.0 g

Limiting reactant = ?

Mass of ammonia formed = ?

Mass of excess reactant left = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 165.0 g/ 28 g/mol

Number of moles = 5.9 mol

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 215.0 g/ 2 g/mol

Number of moles = 107.5 mol

Now we will compare the moles of ammonia with both reactant.

H₂ : NH₃

3 : 2

107.5 : 2/3×107.5 = 71.7 mol

N₂ : NH₃

1 : 2

5.9 : 2/1×5.9 = 11.8 mol

Less number of moles of ammonia are formed by the nitrogen it will act as limiting reactant.

Mass of ammonia formed:

Mass = number of moles × molar mass

Mass = 11.8 mol × 17 g/mol

Mass = 200.6 g

Mass of hydrogen left:

We will compare the moles of hydrogen and nitrogen.

N₂ : H₂

1 : 3

5.9 : 3/1×5.9 = 17.7 mol

Out of 107.5 moles 17.7 moles of hydrogen react with nitrogen.

Number of moles left unreacted = 107.5 - 17.7 mol = 89.8 mol

Mass of hydrogen left:

Mass = number of moles × molar mass

Mass = 89.8 mol × 2 g/mol

Mass = 179.6 g