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elixir [45]
3 years ago
6

For the following reaction, 5.21 grams of nitrogen gas are allowed to react with 5.91 grams of hydrogen gas.

Chemistry
1 answer:
xz_007 [3.2K]3 years ago
7 0

Answer:

Maximum amount of ammonia that can be formed is 6.32 g

Explanation:

The balanced equation is:

N2 + 3H2\rightarrow 2NH3

Moles\ N2 = \frac{5.21g}{28g/mol} =0.186\\\\Moles H2 = \frac{5.21g}{2g/mol} =2.61

N2 is the limiting reactant

Based on the reaction stoichiometry:

1 mole N2 produced 2 moles NH3

therefore, 0.186 mol N2 will yield: 2(0.186) = 0.372 moles NH3

Molar mass NH3 = 17 g/mol

Maximum\ Mass\ NH3 = 0.372 mols*17g/mol=6.32g

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<u>Answer:</u>

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<u>Explanation:</u>

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3 years ago
A generic weak acid with formula HA has a Ka = 2.76 x 10-8. Calculate the Kb for the conjugate base of the acid.
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Answer:

3.62x10⁻⁷ = Kb

Explanation:

The acid equilibrium of a weak acid, HX, is:

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Where Ka = [X⁻] [H₃O⁺] / [HX]

And basic equilibrium of the conjugate base, is:

X⁻ + H₂O ⇄ OH⁻ + HX

Where Kb = [OH⁻] [HX] / [X⁻]

To convert Ka to Kb we must use water equilibrium:

2H₂O ⇄ H₃O⁺ + OH⁻

Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

Thus, we can obtain:

Kw = Ka*Kb

Solving for Kb:

Kw / Ka = Kb

1x10⁻¹⁴ /  2.76x10⁻⁸ =

3.62x10⁻⁷ = Kb

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