If we had 10.9 g of nitrogen and 2 g of hydrogen, how much nitrogen would remain if all the hydrogen was consumed?

Method 1:

Molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol
Molar mass of H₂ = 1.0 × 2 = 2.0 g/mol

N₂ + 3H₂ → 2NH₃
Mole ratio N₂ : H₂ = 1 : 3

No. of moles of H₂ reacted = (2 g) / (2.0 g) = 1 mol
No. of moles of N₂ reacted = (1 mol) × (1/3) = 1/3 mol
Mass of N₂ reacted = (1/3 mol) × (28.0 g/mol) = 9.3 g
Mass of N₂ would remain = (10.9 - 9.3) g = 1.6 g

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Method 2:

(2 g H₂) × (1 mol H₂ / 2.0 g H₂) × (1 mol N₂ / 3 mol H₂) × (28.0 g N₂ / 1 mol N₂) = 9.3 g N₂ reacted

(total 10.9 g N₂) - (9.3 g N₂ reacted) = 1.6 g N₂ remains

The product is ammonia.

N2 + H2 → NH3

To balance the nitrogen, make two ammonias.
N2 + H2 → 2 NH3

To balance the hydrogen, make three of it.
N2 + 3 H2 → 2 NH3

According to the coefficients in the balanced equation, one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia. Let’s determine the number of moles of nitrogen and hydrogen.

For N2, n = 10.9 ÷ 28
This is approximately 0.4 mole.
For H2, n = 2 ÷ 2 = 1 mole

Since all of the hydrogen was consumed, the number of moles of nitrogen that was consumed is one third. To determine the number of moles of nitrogen that was not consumed, subtract one third from 10.9 ÷ 28.

For N2, n = (10.9 ÷ 28) – ⅓

This is approximately 0.0556 mole. To determine the mass of nitrogen that is not consumed, multiply this number by 28.

Mass = 28 * [(10.9 ÷ 28) – ⅓]
This is approximately 1.57 grams. I hope this is helpful for you.