Question

Please i need help urgently please . formula N2+ 3H2=2NH3 . how many moles of ammonia would you get if 4.5 grams of hydrogen gas reacted with N2?

Answer 1

To determine the number of moles of ammonia (NH3) produced when 4.5 grams of hydrogen gas (H2) reacts with nitrogen gas (N2), we need to use the balanced chemical equation and molar masses of the substances involved.

The balanced equation is: N2 + 3H2 → 2NH3

Step 1: Calculate the molar mass of H2:

H2: 2 (atomic mass of hydrogen) = 2 g/mol

Step 2: Convert grams of H2 to moles:

Moles of H2 = mass of H2 / molar mass of H2

Moles of H2 = 4.5 g / 2 g/mol = 2.25 mol

Step 3: Apply the mole ratio from the balanced equation:

From the balanced equation, we see that 3 moles of H2 react to form 2 moles of NH3.

Step 4: Calculate the moles of NH3 produced:

Moles of NH3 = (moles of H2) × (2 moles of NH3 / 3 moles of H2)

Moles of NH3 = 2.25 mol × (2 mol / 3 mol) ≈ 1.5 mol

Therefore, approximately 1.5 moles of ammonia (NH3) would be produced when 4.5 grams of hydrogen gas (H2) reacts with nitrogen gas (N2).

Answer 2

To calculate the number of moles of ammonia (NH3) produced when 4.5 grams of hydrogen gas (H2) reacts with nitrogen gas (N2), we need to use the balanced equation and molar masses of the substances involved.

The balanced equation is:

N2 + 3H2 → 2NH3

The molar mass of H2 is 2 grams/mol.

To find the number of moles of H2, we divide the given mass by its molar mass:

moles of H2 = mass of H2 / molar mass of H2

moles of H2 = 4.5 g / 2 g/mol

moles of H2 = 2.25 mol

According to the balanced equation, the stoichiometric ratio between H2 and NH3 is 3:2. This means that for every 3 moles of H2, we get 2 moles of NH3.

Using this ratio, we can calculate the number of moles of NH3 produced:

moles of NH3 = (moles of H2 / 3) * 2

moles of NH3 = (2.25 mol / 3) * 2

moles of NH3 = 1.5 mol

Therefore, if 4.5 grams of hydrogen gas reacted with nitrogen gas according to the given equation, it would produce 1.5 moles of ammonia.

Step-by-step explanation: