Question

Fritz Haber, a German chemist, discovered a way to synthesize ammonia gas (NH3) by combining hydrogen and nitrogen gases according to the following equation:

3H2(g) + N2(g) → 2NH3(g)

What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?
L
What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
L

Answer 1

To produce 250.0 L of ammonia at STP, 125.0 L of nitrogen is needed. To produce 2.50 mol of ammonia at STP, 84.0 L of hydrogen is required.

Step-by-step explanation:

The equation for the synthesis of ammonia gas (NH3) is given as:

N2(g) + 3H2(g) → 2NH3(g)

To find the volume of nitrogen needed to produce 250.0 L of ammonia gas at STP, we look at the stoichiometric relationship between nitrogen and ammonia, which is 1:2. Therefore, for every 2 volumes of ammonia, we need 1 volume of nitrogen. Thus:

Volume of N2 = (250.0 L NH3) × (1 volume N2 / 2 volumes NH3) = 125.0 L N2

For the volume of hydrogen needed to produce 2.50 mol of ammonia at STP, using the stoichiometric relationship of 3:2 between hydrogen and ammonia, we calculate the moles of hydrogen needed first:

Moles of H2 = 2.50 mol NH3 × (3 mol H2 / 2 mol NH3) = 3.75 mol H2

At STP, 1 mole of gas occupies 22.4 L, so:

Volume of H2 = 3.75 mol × 22.4 L/mol = 84.0 L H2

Answer 2

Volume of nitrogen needed = 125.5 L

Volume of hydrogen needed = 84.05 L

Step-by-step explanation:

Given data:

Volume of ammonia produced = 250.0 L

Volume of nitrogen needed = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Number of moles of ammonia:

PV = nRT

1 atm× 250.0 L = n × 0.0821 amt.L/mol.K × 273K

250.0 atm.L = n × 22.41 amt.L/mol

n = 250.0 atm.L /22.41 amt.L/mol

n = 11.2 mol

Now we will compare the moles of ammonia with nitrogen.

NH₃ : N₂

2 : 1

11.2 : 1/2×11.2 = 5.6 mol

Volume of nitrogen needed:

PV = nRT

1 atm× V = 5.6 mol × 0.0821 amt.L/mol.K × 273K

V = 125.5amt.L/1 atm

V = 125.5 L

2nd)

Given data:

Number of moles of ammonia produced = 2.50 mol

Volume of hydrogen needed = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Now we will compare the moles of ammonia and hydrogen.

NH₃ : H₂

2 : 3

2.50 : 3/2×2.50 = 3.75 mol

Volume of hydrogen:

PV = nRT

1 atm × V = 3.75 mol × 0.0821 amt.L/mol.K × 273K

V = 84.05 amt.L/1 atm

V = 84.05 L