Write a balanced half-reaction for the reduction of gaseous nitrogen (N2) to aqueous hydrazine (N2H4) in basic aqueous solution.

Chemistry

1 answer:

weqwewe [10]4 years ago

5 0

Answer:

N₂(g) + 4H⁺(aq) + 4e⁻ → N₂H₄(aq)

Explanation:

The half reaction for the reaction for the reduction of gaseous nitrogen to aqueous hydrazine is;

N₂(g) → N₂H₄(aq)

The balancing the atoms in the half reaction. Hydrogen atom is balanced by adding hydrogen ions (H⁺)

We have;

N₂(g) + 4H⁺(aq) → N₂H₄(aq)

Then we balance the charge on both sides by adding electrons where the positive charge is greater;

we have;

N₂(g) + 4H⁺(aq) + 4e⁻ → N₂H₄(aq)

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3H2 + N2 —> 2NH3 .

In-s [12.5K]

Answer:

$\boxed{1.26 \times 10^{20} \text{ molecules NH}_{3}}$

Explanation:

We will need a balanced chemical equation with masses, moles, and molar masses.

1. Gather all the information in one place:

M_r: 2.016 17.03

3H₂ + N₂ ⟶ 2NH₃

m/g: 6.33 × 10⁻⁴

2. Calculate the moles of H₂

$\text{Moles of H}_{2} = 6.33 \times 10^{-4}\text{ g H}_{2} \times \dfrac{\text{1 mol H}_{2}}{\text{2.016 g H}_{2}} = 3.140 \times 10^{-4} \text{ mol H}_{2}$

3. Calculate the moles of NH₃

The molar ratio is 2 mol NH₃/3 mol H₂.

$\text{Moles of NH}_{3} = 3.140 \times 10^{-4} \text{ mol H}_{2} \times \dfrac{\text{2 mol NH}_{3}}{\text{3 mol H}_{2}} = 2.093 \times 10^{-4}\text{ mol NH}_{3}$

4. Calculate the molecules of NH₃

There are 6.022 × 10²³ molecules of NH₃/1 mol NH₃.

$\text{Molecules of NH}_{3} = 2.093 \times 10^{-4}\text{ mol NH}_{3} \times \dfrac{6.022 \times 10^{23}\text{ molecules NH}_{3}}{\text{1 mol NH}_{3}}\\= 1.26 \times 10^{20}\text{ molecules NH}_{3}\\\text{The reaction produces } \boxed{\mathbf{1.26 \times 10^{20}} \textbf{ molecules NH}_{3}}$